First of all, if $x:U\subset \mathbb R^2\rightarrow S$ is a parametrization, then $x^{-1}: x(U) \rightarrow \mathbb R^2$ is differentiable: indeed, following the very definition of a differentiable map from a surface, $x$ is a parametrization of the open set $x(U)$ and since $x^{-1}\circ x$ is the identity map, it is differentiable.
Now, let $p$ be a point on the surface $S$, $x:U\subset \mathbb R^2\rightarrow S$ be a parametrization s.t. $x(0)=p$ and $y:V\subset \mathbb R^2\rightarrow S$ be another parametrization s.t. $L(p)=y(0)$.
To make it clear, let's say that $x(u,v)=(x_1(u,v),x_2(u,v),x_3(u,v))$ and $y^{-1}(x,y,z)=(\varphi_1(x,y,z),\varphi_2(x,y,z))$ then the map $L\circ x:U\rightarrow S$ is given by : $$L\circ x (u,v)=\begin{pmatrix} a&b&c\\d&e&f \\g&h&i\end{pmatrix}\begin{pmatrix} x_1(u,v) \\ x_2(u,v) \\ x_3(u,v) \end{pmatrix}$$
So $f(u,v)=y^{-1}\circ L \circ x(u,v)$ looks like $$f(u,v)=y^{-1}\circ L \circ x(u,v)=\\\ \begin{pmatrix}\varphi_1(ax_1(u,v)+bx_2(u,v)+cx_3(u,v),\cdots,gx_1(u,v)+hx_2(u,v)+ix_3(u,v)) \\ \varphi_2(gx_1(u,v)+hx_2(u,v)+ix_3(u,v),\cdots,gx_1(u,v)+hx_2(u,v)+ix_3(u,v))\end{pmatrix}$$
which is clearly differentiable.
Moreover, you can easily check using the chain rule that $$df_0=d(y^{-1})_{L(p)}\circ L \circ dx_0.$$
Roughly speaking, this map does : $$\mathbb R^2 \underset{dx}{\longrightarrow} T_pS \underset{L}{\longrightarrow} T_{L(p)}S\underset{dy^{-1}}{\longrightarrow} \mathbb R^2$$
which means that you send a vector of $\mathbb R^2$ onto $T_pS$ using the parametrization $x$ (it always gives you a good basis of the tangent space), then L acts and you read the information again using the second parametrization $y$ that takes the new vector onto $\mathbb R^2$.
So $L$ is nothing else but the derivative of $L:S\rightarrow S$ as a map between two surfaces.
In fact, this has to be expected because you might know that the derivative of a linear map between two vector spaces does not depend on the point and is equal to itself, so it has to be the same for surface or submanifold in general.
I'm assuming you are working with the definitions given in Do Carmo's "Curves and Surfaces" book and that $V \subseteq \mathbb{R}^3$ is an open set. First, let us write precisely the domains and ranges of the maps involved:
- The map $\mathbf{x}_1 \colon U_1 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_1 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_1(U_1) \subseteq S_1$.
- The map $\varphi \colon U \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U \subseteq \mathbb{R}^3$ is an open set and $\varphi(S_1) \subseteq S_2$.
- The map $\mathbf{x}_2 \colon U_2 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_2 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_2(U_2) \subseteq S_2$.
As you have noted, $\mathbf{x}_2^{-1} \colon \mathbf{x}_2(U_2) \rightarrow U_2$ is not classically differentiable but is differentiable in the sense of a map between a regular surface $\mathbf{x}_2(U_2)$ and an open set in $\mathbb{R}^2$ (if you want, you can embed $\mathbb{R}^2$ in $\mathbb{R}^3$ and consider $U_2$ as a regular surface in the sense of Do Carmo and then $\mathbf{x}_2^{-1}$ will be differentiable in the sense of a map between two regular surfaces but you won't gain much from it).
Thus, you have two options:
- Prove that $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow \mathbf{x}_2(U_2)$ is a differentiable map between an open subset in $\mathbb{R^2}$ and a regular surface. Then, if you already proved this, apply the chain rule for maps between open subsets and regular surfaces to deduce that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is smooth as a map between open subsets (or as a map between regular surfaces - the two notions concide).
- Show that one can extend the map $\mathbf{x}_2^{-1}$ to a map $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ where $\mathbf{x}_2(U_2) \subseteq V$ and $V$ is an open subset of $\mathbb{R}^3$ such that $\hat{\mathbf{x}}_2^{-1}$ is differentiable. Then, use the $\mathbb{R}^n$ version of the chain rule to deduce that the composition of $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow V$ and $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ is smooth, thus showing that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is also smooth. This is a standard argument done using the inverse function theorem and you can see an example of it in page 70 of Do Carmo's book.
Best Answer
$$g_1^{-1}\circ\varphi\circ f_1=g_1^{-1}\circ g_2\circ (g_2^{-1} \circ\varphi\circ f_2)\circ f_2^{-1}\circ f_1= \\=G \circ (g_2^{-1} \circ\varphi\circ f_2)\circ F$$
and also
$$g_2^{-1}\circ\varphi\circ f_2=g_2^{-1}\circ g_1\circ (g_1^{-1} \circ\varphi\circ f_1)\circ f_1^{-1}\circ f_2= \\=G^{-1} \circ (g_1^{-1} \circ\varphi\circ f_1)\circ F^{-1}$$
The maps $F$ and $G$ are diffeomorphisms: they are differentiable, bijective and their inverses are also differentiable.
Now use: a composition of differentiable functions is also differentiable.