What doCarmo might have had in mind:
I didn't check, but I think the quotient map
$$ \pi: C \to M $$
is precisely the orientation covering of $M$. And one can prove that this covering is connected iff $M$ is not orientable (see Lee "Intro to Smooth Manifolds", p. 331 for example).
If the above is not useful: Maybe you could try to pull back the orientation of $M$ to $C$ and get a contradiction or so. (although this really has not much to do with part a) of the exercise, so it might be that doCarmo wants you to imbed the Möbius strip)
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.
Best Answer
Hint : I guess your definition is the following: A regular surface $S\subset \mathbb R^3$ is orientable if one can find a family $(X_a)_{a\in \alpha}$ so that $\det(dX^{-1}_bX_a)) >0$.
So if $S$ is orientable, then you are given that family of charts $X_a$'s. For each chart, one can define locally the normal vector
$$N_a = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}$$
So is it true that you can extends this map to the whole $S$? Or, in another chart $X_b$, do you have $N_a = N_b$? If you can show that, then $N: S\to \mathbb R^3$ is the mapping you want.
On the other hand, if such a $N$ is given, then you can consider all charts $X_a$ of $S$ so that
$$N = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}. $$
Will this families of charts $X_a$'s satisfy $\det(dX^{-1}_bX_a)) >0$?