I've read many questions and answers, but there's this concept I can't understand, in relation with polar coordinates.
In "A Visual Introduction to Differential Forms", the author (Fortney) defines directional derivative as:
Let $f:\mathbb{R}^n \to \mathbb{R}$ be a real-valued function, let $p
\in \mathbb{R}^n$ be a point. let $\mathbf{v}_p$ be a vector tangent
to manifold $\mathbb{R}^n$, that is, $\mathbf{v}_p \in
T_p(\mathbb{R}^n)$. The number
$$ \tag{1} \mathbf{v}_p [f] = \frac{d}{dt} \big( f(p+t \mathbf{v}_p ) \big) \big| _{t=0} $$
is called the directional derivative of $f$ w.r.t. $\mathbf{v}_p$.
He then defines $\mathop{grad}$ of $f$ at $(r_0, \theta_0)$ in polar coordinates:
$$\tag{2} \mathop{grad} f = \frac{\partial f}{\partial r} \hat{\mathbf e}_r + \frac{1}{r_0} \frac{\partial f}{\partial \theta} \hat{\mathbf e} _\theta $$
Let's look at $\hat{\mathbf e}_{\theta}$. I assume it's the vector $ \begin{bmatrix} 0 \\ 1 \end{bmatrix} $, but in what coordinate system? is it when the basis are $\big( \hat{\mathbf e} _r, \hat{\mathbf e} _{\theta} \big)$, or is the basis $\big( \hat{\mathbf e} _r, \frac{1}{r} \hat{\mathbf e} _{\theta} \big)$?
Also, I'm trying to insert $\hat{\mathbf e} _{\theta}$ into the definition of (1), and see what I get. Let $p=(r_0, \theta_0)$. Then $$ \tag{3} \hat{\mathbf e} _{\theta} [f] = \frac{d}{dt} \big( f(p + t \hat{\mathbf e} _{\theta}) \big) \big| _{t=0} $$
I'm trying to visualize what $\hat{\mathbf e} _{\theta}$ means. At some point $p=(r_0, \theta_0)$, we can look at two "curves" - the first is a stright-line (denoted in red), the other is the "circle", denoted in black. What displacement does $\hat{\mathbf e} _{\theta}$ represent - along the red curve or the black curve?
