Given a smooth manifold $M$, how does local coordinates affect the basis of $T_p(M)$? Let's say I have a smooth manifold $M$ of dimension $n$ and a smooth chart $(U, \phi)$ consisting of  open set $U$ of $M$ along with a homeomorphism $\phi  : U \to \widehat{U} =\phi[U] \subseteq \mathbb{R}^n$. 
Recall that the component functions $x^i : U \to \mathbb{R}$ of $\phi$ defined by $\phi(p) = (x^1(p), \dots, x^n(p))$ are called the local coordinates on $U$.
Now the basis for $T_p(M)$ is given by $$\left\{\frac{\partial}{\partial x^1}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right\}$$ where each $\frac{\partial}{\partial x^i}\bigg|_p$ is the derivation defined by $$\frac{\partial}{\partial x^i}\bigg|_p(f) = \frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)  \ \ \ \ \ (*)$$ for $f \in C^{\infty}(U)$.
Now in Introduction to Smooth Manifolds by John Lee, it's said that these tangent vectors $\frac{\partial}{\partial x^i}\bigg|_p$ are called the coordinate vectors at $p$ assosciated with the given coordinate system.
But I don't exactly see how the basis vectors $\frac{\partial}{\partial x^i}\bigg|_p$ depend on a given coordinate system other than simply the homeomorphism $\phi$. To me it seems that the right hand side of $(*)$, that being $\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$, is just the usual partial derivative operator in $\mathbb{R}^n$, so by that I mean that $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$
where $e_i$ is the element $(0, \dots, 1, \dots, 0) \in \mathbb{R}^n$ with a $1$ in the $i$-th position of the $n$-tuple. Am I correct in saying that? I think I am wrong, because what if I have some $2$-dimensional manifold $N$, and a chart $(V, \psi)$ where points of $\widehat{V}$ are best expressed in polar coordiantes and I wanted to compute $\frac{\partial}{\partial x^i}\bigg|_{q}(f)$ for some $q \in N$ and $f \in C^{\infty}(V)$. I don't see how my interpretation of $(*)$ would make sense then there.
Maybe I'm getting lost in notation but for example I don't understand what the relation between the symbolic $x^i$'s in the notation of the component functions $x^i$ of $\phi$, the basis vectors $\frac{\partial}{\partial x^i}\bigg|_{p}$, and $\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) $ are other than they are simply a reminder for the latter two that we're taking the $i$-th partial derivative in $\widehat{U}$. The notation used suggests there must be some stronger relationship between these.
 A: I'm not shure I have fully understood your question, but hopefully what I say can help.

1) When you say it's said that these tangent vectors $\frac{\partial}{\partial x^i}\bigg|_p$ are called the coordinate vectors at $p$ assosciated with the given coordinate system.

Here by the given coordinate system we mean simply the chart $\phi$. If $\phi=(x^1,\dots,x^n)$ then $(x^1,\dots,x^n)$ are exactly the coordinates we are referring to.

2) As regards the symbolic $x^i$'s, they are present in the notation of the basis vectors $\frac{\partial}{\partial x^i}\bigg|_{p}$, beacause this definition depends on $\phi$ and so on the function $x^i$'s, but they should not be in the notation of  $\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) $.

Beacuse, as you sad, $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$ and in the right hand side we have no link to the coordinate functions of $\phi$, i.e. to te $x^i$'s.
Maybe it could be better to write $$\partial_i\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$
Definitely, we should define $\frac{\partial}{\partial x^i}\bigg|_{p}f=\partial_i\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$
