I'm not sure I'll be able to answer all your questions, but I'll have a go, but first some background to make sure we're on the same page. I think the most intuitive way to view tangent vectors is as an equivalence class of smooth curves in the appropriate manifold. So, an element of the tangent bundle $TM$ is given as an equivalence class $[s\mapsto \gamma(s)]$ of smooth curves $\gamma:\Bbb{R} \to M$ (strictly speaking, it is only defined in a small open interval containing $0$). Understanding this in a chart is also pretty simple I think.
Say we model $M$ on an $m$-dimensional vector space $E$ (typically we take it to be $\Bbb{R}^m$), and that $(U,\alpha)$ is a chart on $M$; then we can construct a chart $(TU,T\alpha)$ for the tangent bundle as
\begin{align}
T\alpha: TU &\to \alpha[U]\times E\\\\
T\alpha([\gamma]) &:= \left((\alpha\circ \gamma)(0), (\alpha \circ \gamma)'(0)\right)
\end{align}
In words: we have an equivalence class of curves $[\gamma]$ in the manifold $M$. So, $\alpha \circ \gamma$ is a curve in the vector space $E$. What we do is map $[\gamma]$ to its chart-representative base point and the chart-representative velocity vector.
Which roles do the vector component $\dot{q} \in T_{\pmb q}M$ and the fiber change component $\dot{q} \in TTM$ play generally in manifolds?
Since $T^2M$ is defined as $T(TM)$, it means all we have to do is carefully reapply the definitions again. Now, an element of the second tangent bundle is by definition an equivalence class $[t\mapsto \Gamma(t)]$ of smooth curves into the tangent bundle $\Gamma:\Bbb{R}\to TM$. So, $\Gamma(t) \in TM$, which means it is itself an equivalence class of curves in $M$, say $\Gamma(t) = [s\mapsto \gamma(s,t)]$. So, we have
\begin{align}
[t\mapsto \Gamma(t)] &= \left[t\mapsto [s\mapsto \gamma(s,t)]\right] \in TTM
\end{align}
This is what a general element of the second tangent bundle looks like. Once again, we can consider the associated chart $(T^2U, T^2\alpha)$ on the second tangent bundle, and considering how things look in a chart should help to clarify the different roles played by the two $\dot{q}$ you wrote in your post (which tbh I think is an abuse of notation). Ok, so let's calculate:
\begin{align}
T^2\alpha : T^2U \to (\alpha[U] \times E) \times (E\times E)
\end{align}
is given by
\begin{align}
T^2\alpha([\Gamma]) &:= \left((T\alpha \circ \Gamma)(0), (T\alpha \circ \Gamma)'(0)\right) \tag{$*$}
\end{align}
To proceed further, we should understand what $(T\alpha \circ \Gamma)(t)$ looks like so that we can evaluate at $t$ and evaluate its derivative at $t=0$. By definition,
\begin{align}
(T\alpha \circ \Gamma)(t) &= T\alpha([s\mapsto \gamma(s,t)])\\
&:= \left((\alpha\circ \gamma)(0,t), \dfrac{d}{ds}\bigg|_{s=0}(\alpha\circ \gamma)(s,t)\right) \tag{$**$}
\end{align}
So, by plugging $(**)$ into $(*)$, we find that (removing a few brackets)
\begin{align}
T^2\alpha([\Gamma]) &=\left((\alpha\circ \gamma)(0,0), \dfrac{d}{ds}\bigg|_{s=0}(\alpha\circ \gamma)(s,0),\dfrac{d}{dt}\bigg|_{t=0}(\alpha\circ \gamma)(0,t),\dfrac{d}{dt}\bigg|_{t=0}\dfrac{d}{ds}\bigg|_{s=0}(\alpha\circ \gamma)(s,t)\right) \\\\
&= \left((\alpha\circ \gamma)(0,0),
\dfrac{\partial (\alpha\circ \gamma)}{\partial s}\bigg|_{(0,0)},
\dfrac{\partial (\alpha\circ \gamma)}{\partial t}\bigg|_{(0,0)},
\dfrac{\partial^2 (\alpha\circ \gamma)}{\partial t\partial s}\bigg|_{(0,0)}\right) \tag{$\ddot{\smile}$}
\end{align}
For the sake of visualization, I like to imagine a doubly parametrized curve $(s,t)\mapsto \gamma(s,t)$ from $\Bbb{R}^2 \to M$, so that its image forms a sort of "2-D grid" on the manifold $M$ (for example, think of $M = \Bbb{R}^2$ and $\gamma(s,t) = (s,t)$; then the image of $\gamma$ is exactly the "coordinate grids" on the plane). So you see although there are two first derivatives appearing above, they are taken with respect to the different variables $s$ and $t$; i.e we're differentiating the doubly-parametrized curve along different directions. This should hopefully make it clear that the two $\dot{q}$ you wrote is an abuse of notation, because they're capturing first-order changes but in "different directions".
Which role do they play in the Hamilton formalism (if any different)?
The answer to your question as stated is that they do not play a role, because Hamiltonian mechanics happens on $T^*M$ (if $M$ is the configuration space). But I think alot of your confusion comes from naively placing dots over the various things. To me it is much clearer to speak of an equivalence class of curves $[s\mapsto \gamma(s)]$ when speaking of a tangent vector in M, because in this notation, the base point is clear; it is $\gamma(0)$. So $[s\mapsto \gamma(s)] \in T_{\gamma(0)}M$. If we go to the second tangent bundle then the general element looks like $[t\mapsto [s\mapsto \gamma(s,t)]]$, and this lies in the specific fiber $[t\mapsto [s\mapsto \gamma(s,t)]]\in T_{[s\mapsto \gamma(s,0)]}(TM)$.
Just for fun, lets see how things look in the third tangent bundle: $[t_3, \mapsto [t_2 \mapsto [t_1 \mapsto \gamma(t_1,t_2,t_3)]]]$ is a general element of $TTTM$, and it happens to lie in the specific fiber $T_{[t_2\mapsto [t_1\mapsto \gamma(t_1,t_2,0)]]}(TTM)$. i.e we just set the "outermost parameter" to $0$, and thats the base point. So, in general on $T^nM$, the general element looks like $[t_n \mapsto \dots [t_1\mapsto \gamma(t_1, \dots, t_n)]]$, and the base point for this is $[t_{n-1}\mapsto \dots [t_1 \mapsto \gamma(t_1, \dots, t_{n-1}, 0)]] \in T^{n-1}M$
How to construct a second derivative $f_{* *}: TTM \to TTN$? How do these components appear there?
Well you just do it recursively. $f_{**}$ is by definition $(f_*)_*$. So, if $[t\mapsto [s\mapsto \gamma(s,t)]] \in TTM$, then
\begin{align}
f_{**}([t\mapsto [s\mapsto \gamma(s,t)]]) &:= [t\mapsto f_*([s\mapsto \gamma(s,t)])]\\
&:=[t\mapsto [s\mapsto (f\circ\gamma)(s,t)]]
\end{align}
So, if we take a chart $(V,\beta)$ on the target manifold $N$, and we lift it to a chart $(T^2V, T^2\beta)$, then this looks like (by a simple application of $(\ddot{\smile})$)
\begin{align}
\left((\beta\circ f\circ \gamma)(0,0),
\dfrac{\partial(\beta\circ f\circ \gamma)}{\partial s}\bigg|_{(0,0)},
\dfrac{\partial(\beta\circ f\circ \gamma)}{\partial t}\bigg|_{(0,0)},
\dfrac{\partial^2(\beta\circ f\circ \gamma)}{\partial t\partial s}\bigg|_{(0,0)}\right)
\end{align}
- How, if at all, does this relate to curvature and torsion of curves?
- How, if at all, does the exterior derivative $dd=0$ or any other relevant derivative relate to this?
I'm not sure, I don't see any obvious links... nothing comes to mind right now.
- The Hamilton equations of motion are $\dot{\pmb q} = \frac{\partial H}{\partial \pmb p}, \dot{\pmb p} = -\frac{\partial H}{\partial \pmb q}$ (with $H: T^*M\to\mathbb R$ and $(\pmb q, \pmb p) \in T^*M$). How does the notational double occupancy of $\dot{\pmb q}$ resolve here?
I don't see a double usage of $\dot{\pmb q}$ here. One thing to note however is that the symbols $q,\dot{q}, p$ are often used to mean different things in different contexts, so we abuse notation quite often. For example, if I have a manifold $M$ with a chart $(U,\alpha)$, where $\alpha:U \to\alpha[U]\subset \Bbb{R}^n$, then we often define $q^i := \text{pr}^i_{\Bbb{R}^n}\circ \alpha$; this is now a function $U \to \Bbb{R}$. Next, if we consider the chart $(TU, T\alpha)$ on $TM$ then note that $T\alpha$ is a map $TU \to \alpha[U] \times \Bbb{R}^n \subset \Bbb{R}^n \times \Bbb{R}^n$. So, we can consider the 2n coordinate functions $\text{pr}^{\mu}_{\Bbb{R}^{2n}}\circ T\alpha$, where $\mu\in \{1,\dots, 2n\}$. Then, you can prove that for $i\in \{1,\dots, n\}$, $\text{pr}^{i}_{\Bbb{R}^{2n}}\circ T\alpha = q^{i}\circ \pi_{TM}$, where $\pi_{TM}:TM \to M$ is the standard bundle projection. However, people usually omit the composition by $\pi_{TM}$ in the notation, so that $q^{i}$ can stand for either a function on $U$ or $TU$.
If you now introduce the cotangent bundle into the mix, then $q^i$ has $3$ different meanings, either $q^i$ or $q^i \circ \pi_{TM}$ or $q^i \circ \pi_{T^*M}$; which meaning is intended should be decided based on context. Anyway here's an answer I wrote regarding how to interpret Hamilton's equations (though you should note that what you're calling $M$ here is what the OP there calls $Q$, and your $T^*M$ is his $T^*Q = M$).
- In this question I concentrated on $TTM$ while the Hamilton formalism is defined on $TT^*M$. Is there a fundamental difference between $TT^*M$ and $TTM$ that is relevant to the problem in question?
Perhaps this is just a difference in how we want to say things, but I'd say that Hamiltonian mechanics is formulated on $T^*M$, not $T(T^*M)$. Because the hamiltonian is a function $H:T^*M \to \Bbb{R}$, the symplectic form $\omega$ is a $2$-form defined on $T^*M$. All functions, vector fields and forms are defined on $T^*M$.
So, I'm not sure there's any relevance between $TT^*M$ and $TTM$ which is relevant... though you might be interested to know that given a Lagrangian, i.e a function $L:TM \to \Bbb{R}$, we can consider the fiber derivative $FL:TM \to T^*M$, and if this map is a diffeomorphism, I think it allows us to go back and forth between the Lagrangian and Hamiltonian pictures.
In any case, it is a good idea to work out in detail, starting from a chart $(U,\alpha)$ on $M$, what the charts $(TTU, TT\alpha)$ on $TTM$ looks like (of course I already did this above), and also what the chart $(TT^*U, TT^*\alpha)$ on $TT^*M$ looks like. Part of this exercise is to give precise definitions for these charts as well. Of course, my sugestion to you is to do things in the simple case first: what do the charts on the tangent and cotangent bundle $TM$ and $T^*M$ look like?
Now you just have to proceed recursively, and carefully apply definitions. If at any point you get confused, just introduce new letters, say $N:= TM$ and $S:= T^*M$, and then ask yourself what do charts on $TN$ and $TS$ look like. This is of course slightly confusing, and takes some getting used to, but I think there's no way around it.
I realize I may not have answered your questions to complete satisfaction, but hopefully this is enough to get you started.
