If $S_1,S_2$ are regular surfaces and $\varphi:U\subset S_1\rightarrow S_2$ is a differentiable map such that the differential $d\varphi_p$ of $\varphi$ at $p\in U$ is an isomorphism, then $\varphi$ is a local diffeomorphism at $p$.
Suppose $X_1:V_1\subset \mathbb{R}^2\rightarrow S_1 $, $X_2:V_2\subset \mathbb{R}^2\rightarrow S_2$ are parametrizations. Then $\varphi$ is diffentiable reads $h=X_2^{-1}\varphi X_1:V_1\rightarrow V_2$ is differentiable. Let $q\in V_1$ such that $X_1(q)=p$. By the chain rule we have $d(X_2^{-1}\varphi X_1)_q=dX_2^{-1}{_{\varphi(p)}}\circ d\varphi_p \circ dX_1{_q}$
Since $d\varphi_p$ is a isomorphism, it is invertible. Then by the inverse function theorem, there exists $q\in W_1\subset V_1$ and $h(q)\in W_2\subset V_2$. Such that $h:W_1\rightarrow W_2$ has its inverse $h^{-1}:W_2\rightarrow W_1$.
My question is: if we can say $h^{-1}=X_2\psi X_1^{-1}$ and $\psi $ is the differentiable inverse of $\varphi$?
No, you mean, of course, $h^{-1} = (X_2^{-1}\varphi X_1)^{-1} = X_1^{-1}\varphi^{-1}X_2$. If you draw a diagram of the mappings, this will fall out. Your composition of functions actually is not defined.