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.
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$_2$.
Best Answer
The formal details involve taking the usual sense of "diffeomorphism" between open sets and fitting it to the more gentle definition to manifolds. In order to prove $\phi|_S$ is a diffeomorphism, we need to show the following
1) $\phi|_S$ is differentiable: Let $\psi:U\rightarrow S$ be a paremeterzation of $S$. The composition $\phi \circ \psi: U\rightarrow O_2$ is differentiable in the "usual sense" of open sets due to the chain rule, as both $\phi,\psi$ are differentiable on open sets. This now follows from definitions A,B.
2) $\phi|_S$ is invertible with differentiable inverse: We know $\phi |_{O_1}$ is bijective, which must mean $\phi|_S$ is also bijective as a restriction. We know there exists $\phi^{-1}:\phi(O_1) \rightarrow O_1$ differentiable (between open sets). Let $\varphi$ be a parameterization of $\phi(S)$. Just like in (1), $\phi^{-1}\circ \varphi$ is differentiable and therefore from defition A, $\phi^{-1}$ is differentiable in the "surface sense"
I don't know how diffeomorphisms are defined in this book, so you might have to replace every "differentiable" in my answer with $C^k$.
Another note: I should point out $\phi(O_1)$ is open because $\text{rank}D\phi = 3$ which by the open mapping theorem, means $\phi$ is an open mapping. (This can be also proved easily by the inverse function theorem though. Let me know if you need me to elaborate and fill in the details!).