Conceptually it is clear what Claudio Arezzo is saying in his proof that the derivative of a function $f: S\to \mathbb R^n $ at a point $p \in S$ is well defined, i.e. it doesn't depend on $t,$ even if the definition of $df_p(\vec v)=\frac d{dt}\vert_{t=0}(f\circ \alpha)(t)=(f\circ \alpha)'(0)$ includes the idea of a curve on $S$ parameterized by $t.$
The "trick" is to express the curve $\alpha(t)$ as $\vec X \circ (\vec X^{-1} \circ \alpha )$ with $\vec X$ being a local chart from $U\to S,$ such that $\vec X_q=p$ and then apply the chain rule as:
$$d\vec X_q\left( \vec X^{-1} \circ \alpha\right)'(0)=\vec v$$
I am trying to reconcile the chain rule $(f\circ g)'=(f'\circ g)\cdot g'$ with this equation, and if $f=\vec X$ and $g=\vec X^{-1} \circ \alpha,$ I am missing an additional term perhaps corresponding to a second application of the chain rule to the composition $\vec X^{-1} \circ \alpha.$
No, you're not missing anything. $q=(X^{-1}\circ\alpha)(0)$ since $X(q)=p=\alpha(0)$. So $dX_q$ is the "$(f'\circ g)$" part of your chain rule expression, and applying to $(X^{-1}\circ\alpha)'(0)$ is the "multiplication by $g'$" part.