There is a similar question to this on stackexchange, but it uses the fact that a change of parameters is a diffeomorphism. The question is essentially about two different forms of differentiability, and how they coincide. I will use two different words for these two senses, to keep them separate. I will use the subscript 1 for differentiability of maps between surfaces, and the subscript 2 for the usual sense, i.e. existence of partial derivatives.
We say that a map $\phi$ between two regular surfaces $S_1$ and $S_2$ is differentiable$_1$ if there exists local parameterizations $X_1$ of $S_1$ and $X_2$ of $S_2$ such that $X_2^{-1}\circ\phi\circ X_1$ is differentiable$_2$. What I want to show is that this definition is independent of parametrization. I know that if a surface is regular, with parametrization $X$, then $X^{-1}$ is differentiable$_1$.
To that end, let $Y_1$ and $Y_2$ be two other parameterizations of $S_1$ and $S_2$, respectively. Then we have
$$
Y_2^{-1}\circ\phi\circ Y_1 = Y_2^{-1}\circ X_2\circ X_2^{-1}\circ\phi\circ X_1 \circ X_1^{-1} \circ Y_1 = (Y_2^{-1}\circ X_2) \circ (X_2^{-1}\circ\phi\circ X_1) \circ (X_1^{-1} \circ Y_1).
$$
The middle term is differentiable$_2$, however, the edge terms consists of compositions of mappings that are differentiable in two different ways. E.g. $X_2$ is differentiable$_2$, but $Y_2^{-1}$ is differentiable$_1$. I don't know how to bring together these two separate sorts of differentiability.
Best Answer
Your worry is legitimate and yes, we do not argue that $X_1^{-1}\circ Y_1$ is differentiable$_2$ using that $X_1^{-1}$ is differentiable$_1$. Indeed, we have not yet proved the well-definedness of differentiability$_1$.
Instead we argue directly.
Claim: Let $X_1 : U \to S_1$ be a regular parametrization. Then for each $p\in U$, there is an open neighborhood $W$ of $X_1(p)$ in $\mathbb R^n$ and a smooth function $\varphi : W\to U$ so that $$\varphi|_{X_1(U)\cap W} = X_1^{-1}.$$ (that is, $X_1^{-1}$ can be extended to a smooth map $\varphi$ locally in the ambient space $\mathbb R^n$).
Proof of claim Let $v_3, \cdots, v_n$ be $n-2$ fixed vector in $\mathbb R^n$ so that $$\tag{1} \left\{ \frac{\partial X_1}{\partial x}(p), \frac{\partial X_1}{\partial y}(p), v_3, \cdots, v_n\right\}$$ is linearly independent. Define $F : U \times \mathbb R^{n-2} \to \mathbb R^n$ by $$ F(x, y, t_3, \cdots, t_n) = X(x, y) + t_3 v_3 + \cdots + t_n v_n.$$
Then at $(p, 0)$, $DF$ is invertible by (1). By the inverse function theorem, there is $V\in U \times \mathbb R^{n-2}$ containing $(p,0)$ and $W$ containing $F(p,0) = X_1(p)$ so that $F : V\to W$ is invertible and has a smooth inverse $\Phi : W\to V$. Then define $\varphi = \Pi \Phi$, where $\Pi : U \times \mathbb R^{n-2} \to U$ is the projection.
As a simple corollary, $X_1^{-1} \circ Y_1 = \varphi \circ Y_1$ is differentiable$_2$.