From Do Carmo's Differential Geometry of Curves and Surfaces, 2nd edition,
Now, when I tried to prove this, I reached this point:
Up until this box, I couldn't understand the philosophy behind these steps, the definition of $V_{\alpha}$ and $V_{\beta}$ containing all these inverses, compositions, and intersections made me very dizzy, I've tried thinking about them for a good while and I'm still lost, if anyone could guide me through the logic behind these steps I'd really appreciate it, thank you.
Let's try to get some intuition by starting with an easier problem: Let $S_2$ be an orientable regular surface and $\varphi : S_1 \to S_2$ be a (global) diffeomorphism. Prove $S_1$ is orientable.
In this case, we know $S_2$ is a regular surface, so it is covered by a family of coordinate charts $\{(U_\alpha,\mathbf x_\alpha)\}$ each $\mathbf x_\alpha : U_\alpha \subset \mathbb{R}^2 \to S_2$. If two charts overlap, $p \in \mathbf x_\alpha (U_\alpha) \cap \mathbf x_\beta (U_\beta)$, we can write the transition function $\mathbf x_\beta^{-1} \circ \mathbf x_\alpha : \mathbb{R}^2 \to \mathbb{R}^2$. Then since $S_2$ is orientable, we have $\text{det}(d(\mathbf x_\beta^{-1} \circ \mathbf x_\alpha)) > 0$.
Assuming that $\varphi$ is a (global) diffeomorphism, we can cover $S_1$ by a family of coordinate charts $\{\big(U_\alpha,(\varphi^{-1} \circ \mathbf x_\alpha)\big)\}$ each $\varphi^{-1} \circ \mathbf x_\alpha : U_\alpha \subset\mathbb{R}^2 \to S_1$. If two charts overlap, $q \in (\varphi^{-1} \circ \mathbf x_\alpha) (U_\alpha) \cap (\varphi^{-1} \circ \mathbf x_\beta) (U_\beta)$, we can calculate the transition function: $$ (\varphi^{-1} \circ \mathbf x_\beta)^{-1} \circ (\varphi^{-1} \circ \mathbf x_\alpha) = \mathbf x_\beta\circ \varphi \circ \varphi^{-1} \circ \mathbf x_\alpha = (\mathbf x_\beta^{-1} \circ \mathbf x_\alpha)$$
So it is clear that $S_1$ is orientable.
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Returning to the question you posted, the difference is that $\varphi$ is only a local diffeomorphism, which do Carmo defines on page 89. The logic behind the solution you posted is that the intersection $(\varphi^{-1} \circ \mathbf x_\alpha) (U_\alpha) \cap (\varphi^{-1} \circ \mathbf x_\beta) (U_\beta)\subset S_1$ might not be an open set on which $\varphi$ is a diffeomorphism. What we can find instead is for any point in the intersection is a neighborhood $V\subset S_1$ on which $\varphi$ is a diffeomorphism on its image. Then the intersection $V \cap (\varphi^{-1} \circ \mathbf x_\alpha) (U_\alpha) \cap (\varphi^{-1} \circ \mathbf x_\beta) (U_\beta)$ is in the intersection of two charts and also $\varphi$ is a diffeomorphism, letting us use that $\varphi^{-1}$ is well-defined and differentiable.
The problem with the solution you posted is that it presumes $\varphi^{-1} : S_2 \to S_1$ is well-defined on every $\mathbf x_\alpha(U_\alpha)$. This is cannot be assumed true, so the family $\varphi^{-1} \circ \mathbf x_\alpha : U_\alpha\to S_1$ might not be well-defined (An example of when this fails is the coordinate chart & local diffeomorphism $\varphi(u,v) = (\sin u \cos v,\cos u\cos v,\sin v)$ on all of $\mathbb{R}$ to the unit sphere). This might be easier to solve by using unit normal vectors on $S_2$ (pg 106-110) together with the characterization that $d\varphi$ is full rank (pg 89).