If I understand correctly, given a function $f$ on a manifold, there is an 'isomorphism' between a vector $\vec{v} \in T_pM$ and a directional dervative $D_vf(p)$.
For example, the change in $f$ in the direction $\vec{v} = 2e_x + 3e_y$ is given by $D_vf(p) = 2\frac{\partial f}{\partial x} + 3\frac{\partial f}{\partial y}\biggr\rvert_{p}$. So it follows that if you assume $e_x \equiv \frac{\partial}{\partial x}$ and $e_y \equiv \frac{\partial}{\partial y}$, then it must be the case that $\vec{v} = D_v(*)$. Reasoning that $e_x \equiv \frac{\partial}{\partial x}$ is discussed in the next two sections.
Extrinsic Step:
Given a curve $r(\theta) = (\cos(\theta), \sin(\theta))$, I understand that $\frac{dr}{d\theta}(\theta) = -\sin(\theta) e_x + \cos(\theta) e_y$ is a tangent vector.
So extrinsically, you can define the $e_\theta$ basis vector of a $S^1$ as $e_{\theta}(\theta) \equiv \frac{d}{d\theta}(r (\theta))$.
The last step will justify getting rid of the $r$.
Intrinsic Step:
I've always thought of $\frac{d}{dx}(x) = \frac{dx}{dx} = 1$. But perhaps this has mislead me, as it causes me to think of a straight number line.
Suppose we are on $R^1$ at $x=2$, then $\frac{dx}{dx}$ can be thought of a vector that whose tail is at $2$ and whose head is at $3$. But if we were to curve $R^1$ so it's bent downward from the origin, then the physical arrow that points from $2$ in the direction of $3$ denoted by $\frac{dx}{dx}$ would also have to be geometrically slanted downwards. So the geometric interpretation of $\frac{d}{dx}(x)$ would actually vary depending on x, and the statement $\frac{dx}{dx} = 1$ is omitting the curvature of the line.
In this case $\frac{d}{dx}(x)$ points in the direction of increasing x, so we have that $e_x(x) \equiv \frac{\partial}{\partial x}(x)$ and therefore $\vec{v} = D_v(*)$.
I'm not sure how to formalize this, or give a concrete example.
I've spent the last 2 days trying to convince myself that derivative operators are literal tangent vectors from SE and various PDFs, and I feel like I finally got an answer reasoning that $\frac{dx}{dx}$ is not constant extrinsically, but I'm not sure if this is/isn't true.
Is this a correct understanding?
Thanks.