For general values of $\alpha$, each coordinate of $F$ wraps $\Bbb R$ around $S^1$ (if $\alpha=0$, then $F$ is just a circle "parallel" to the first "axis"). So, if we imagine a point going to the right in $\Bbb R$ and its respective image point on the torus $S^1\times S^1$, this means that each coordinate (the projection on each of the "axes" $S^1$) keeps spinning around. So, intuitively, we can hope the image $F(\Bbb R)$ to be something like in the figure below.
In fact, we are being optimistic. In general, $F$ is not a closed (i.e. periodic) curve as in the figure. Maybe $F$ keeps "infinitely" going around $S^1\times S^1$ and in this case $F(\Bbb R)$ is dense in $S^1\times S^1$. The fact is that $F$ is a closed curve if and only if $\alpha \in \Bbb Q$. But let's answer the first question first:
For which values of $\alpha$ the map $F$ is open?
Answer: for none of them. Indeed, by the coordinates of $F$ we see that $F$ is continuous ($C^\infty$, in fact). Suppose that for some value of $\alpha$ the map $F$ is also open. Suppose also that $\alpha\neq 0$ (the case $\alpha=0$ is similar). Put $\epsilon=\frac{1}{2}\min\{1,\,1/|\alpha|\}$>0. Then $G:(-\epsilon,\epsilon)\to F(-\epsilon,\epsilon)$, with $G=F|_{(-\epsilon,\epsilon)}$ is open, bijective and continuous, therefore, a homeomorphism. But $(-\epsilon,\epsilon)$ has dimension 1, while $F(-\epsilon,\epsilon)$ has dimension 2 (because it is open in $S^1\times S^1$). This is impossible (because of this). This proves that $F$ is never open.
Comment: I am doubtful about this question. Shouldn't it be: "For which values of $\alpha$ the map $F$ is open onto its image?" Then this would be very related to the cases below. I think $F$ is open onto its image if and only if $\alpha\in \Bbb Q$...
For which values of $\alpha$ the set $F(\Bbb R)$ is a submanifold?
Answer: for and only for rational values of $\alpha$. Let's prove that.
Suppose $\alpha\in \Bbb Q$: suppose also $\alpha> 0$, $\alpha\neq 1$, the cases $\alpha\leq 0$ or $\alpha=1$ are similar. Denote $\alpha=\frac{p}{q}$, with $p,q\in \Bbb N$ having no common factors other than $\pm 1$. Then
\begin{align}F(t+q)&=((\cos 2\pi(t+q),\sin 2\pi(t+q)),(\cos 2\pi\alpha(t+q),\sin 2\pi\alpha(t+q)))\\
&=((\cos (2\pi t+2\pi q),\sin (2\pi t+2\pi q)),(\cos (2\pi\alpha t+2\pi p),\sin (2\pi\alpha t+2\pi p)))\\
&=((\cos 2\pi t,\sin 2\pi t),(\cos 2\pi\alpha t,\sin 2\pi\alpha t))\\
&=F(t),
\end{align}
so $F$ is $q$-periodic. Therefore, parametrizing $S^1$ by $(\cos 2\pi \theta,\sin 2\pi \theta),\,\, \theta \in [0,1]$, we define
\begin{align}
\begin{matrix}
G:&S^1&\to&S^1\times S^1\\
&(\cos 2\pi \theta,\sin 2\pi\theta)&\mapsto&((\cos 2\pi q \theta,\sin 2\pi q\theta),\,(\cos 2\pi p \theta,\sin 2\pi p\theta)).
\end{matrix}
\end{align}
One verifies that $G$ is an immersion from $S^1$ into $S^1\times S^1$. Also, because $p$ and $q$ have no common factors, $G$ is injective. Since $S^1$ is compact and $G(S^1)\subset S^1\times S^1$ is Hausdorff, we have that $G:S^1\to S^1\times S^1$ is a homeomorphism onto its image $G(S^1)$ (by this result). Therefore, $G$ is an embedding or, in other words, $G(S^1)$ is a submanifold of $S^1\times S^1$. So, we are done if we prove that $G(S^1)=F(\Bbb R)$.
Note that $\theta\in [0,1]$, then putting $t=q\theta$, we have \begin{align} G(S^1)&\ni ((\cos 2\pi q \theta,\sin 2\pi q\theta),\,(\cos 2\pi p \theta,\sin 2\pi p\theta))\\
&=((\cos 2\pi t,\sin 2\pi t),\,(\cos 2\pi \alpha t,\sin 2\pi \alpha t))\in F(\Bbb R).
\end{align}
Therefore, $G(S^1)\subset F(\Bbb R)$. Conversely, if $t\in \Bbb R$ is arbitrary, then because $F$ is $q$-periodic, there exists some $t'\in [0,q]$ such that $F(t)=F(t')$. Then, putting $\theta=t'/q\in [0,1]$, by the same equation above with $t'$ replaced for $t$ we get that $F(\Bbb R)\ni F(t)=F(t')=G(\theta)\in G(S^1)$. This shows that $F(\Bbb R)=G(S^1)$ is a submanifold of $S^1\times S^1$.
Suppose now $\alpha\notin \Bbb Q$: I will prove that $F(\Bbb R)$ is not submanifold a bit more intuitively than I has been doing in the item above (it's pretty geometric but somewhat troublesome to write down; maybe I can improve with details later). If $\alpha\notin \Bbb Q$, then $F$ wraps $\Bbb R$ around $S^1\times S^1$ just like the figure above, BUT never closing (i.e. with no period at all). It keeps going on and on, always going spinning around $S^1\times S^1$, in such a way that $F(\Bbb R)$ is dense in $S^1\times S^1$. Therefore, for any point $p\in F(\Bbb R)$, every sufficiently small open neighborhood of $p$ in $F(\Bbb R)$ has infinitely many path-connected components (remember, an open neighborhood of $p$ in $F(\Bbb R)$ is some $U\cap F(\Bbb R)$, for some $U\ni p$ open in $S^1\times S^1$). Therefore, $F(\Bbb R)$ with its induced topology from $S^1\times S^1$ is not locally path-connected. Since every submanifold is itself a manifold, $F(\Bbb R)$ cannot be a submanifold because every manifold is locally path-connected.
