Why are tangent vectors coordinate-dependent? Why does the coordinate basis for $T_pM$ depend on the coordinate chart we are using? Any two charts containing $p$ agree on some neighborhood of $p$, so shouldn't we be able to find a basis for $T_pM$ that is coordinate-independent? For example, let $M$ be a smooth manifold and let $p \in M$ that is contained in two charts $(U, \varphi)$, $(V, \psi)$. Let $\{\frac{\partial}{\partial x^i}|_p\}$ be a coordinate basis for $T_pM$ with respect to $\varphi$. A basis vector acts on $f \in C^{\infty}M$ by $$\frac{\partial}{\partial x^i}|_p f = \frac{\partial (f \circ \varphi^{-1})}{\partial x^i}(p)$$ Similarly, if $\{\frac{\partial}{\partial \tilde{x}^i}|_p\}$ is a coordinate basis for $T_pM$ with respect to to $\psi$ then $$\frac{\partial}{\partial \tilde{x}^i}|_p (f) = \frac{\partial (f \circ \psi^{-1})}{\partial \tilde{x}^i}(p)$$ But the behavior $\varphi^{-1}$ and $\psi^{-1}$ both map to the same neighborhood of $p$ (the neighborhood can be assumed to be small enough). So why isn't $$\frac{\partial (f \circ \varphi)}{\partial x^i}(p) = \frac{\partial (f \circ \psi)}{\partial \tilde{x}^i}(p)$$
 A: Just because charts give some basis, why are you expecting it to be invariant? Consider say in $\Bbb R^2$ two independent vectors. This always give a basis for the tangent space at "any" point, but certainly differentiating a function in different directions gives different values. A coordinate choice gives $n$ directions to differentiate, but different choices give different values. 
A: The answer is in your question! You asked, "So why isn't
$$\frac{\partial (f \circ \varphi)}{\partial x^i}(p) = \frac{\partial (f \circ \psi)}{\partial \tilde{x}^i}(p)?  \ \ (*)$$"
The answer is that they ARE equal. Because, in the definition of the tangent space, we equate vectors $v$ in one coordinate, with $w$ in another local coordinates in $R^n$ exactly when one is the image of the differntial of the change of coordinate function, (and not when they are equal in the usual sense) i.e.
$$D(\varphi \circ \psi ^{-1})(v)= w $$
From this it follows that (*) says that we have the same tangent vector up there on the manifold.
$\varphi \circ \psi ^{-1}$ : change of coordinates from an open subset of $R^n$ to another
$v$: vector with initial position at, say q
$w$ : vector with initial position at $\hat{q}=\varphi \circ \psi ^{-1}(q)$, and equal to $D(\varphi \circ \psi ^{-1})(v).$
Note: Think of the open unit ball in $R^2$ as a manifold. Now, a chart of M, can be the identity from open ball to M. Another chart can be the rotation of the open ball around $0$, say 45 degrees. So, the tangent vector $\vec{e_1}$ at origin in the first coordinate is equal to the vector ${(1/sqrt(2),1/sqrt(2)}$ at origin, and we must equate them since they must define the same tangent on M.
