Proposition $8.3$ - Fundamental groups and covering spaces by Elon Lages Lima Preliminaries maybe important:
Let $M$ and $N$ be oriented manifolds with the same dimension and $f: M \longrightarrow N$ a local diffeomorphism. We say that $f$ is positive (with respect to the chosen orientations) when, for each $x \in M$, the linear isomorphism $f'(x): T_xM \longrightarrow T_{f(x)}N$ is positive (this means $f'(x)$ is orientation preserving, i.e., maps positive basis of $T_xM$ into positive basis of $T_{f(x)}N$).

Proposition $8.2.$ Let $f: M \longrightarrow N$ be a surjective local diffeomorphism, defined on a connected oriented manifold. In order that $N$ be orientable, it is necessary and sufficient that, for any $x,y \in M$ with $f(x) = f(y)$, the linear isomorphism $f'(y)^{-1} \circ f'(x): T_xM \longrightarrow T_yM$ be positive.

My doubt is concerning to the following result:

Proposition $8.3.$ Let $M$ be a connected manifold of class $\mathcal{C}^k$ and $G$ be a properly discontinuous group of diffeomorphisms of class $\mathcal{C}^k$ in $M$. If the quotient space $M/G$ is Hausdorff, then there exists a unique manifold structure of class $\mathcal{C}^k$ in $M/G$ such that the quotient map $\pi: M \longrightarrow M/G$ is a local diffeomorphism of class $\mathcal{C}^k$. Suppose that $M$ is oriented. In order that $M/G$ be orientable, its necessary and sufficient that each diffeomorphism belonging to $G$ preserve orientation.

I'm trying understand the final argument of the proof:

Suppose now that $M$ is oriented and each $\alpha \in G$ is a positive diffeomorphism of $M$. Then the local diffeomorphism $\pi: M \longrightarrow M/G$ satisfies $\pi(x) = \pi(y) \Rightarrow y = \alpha(x)$, with $\alpha \in G$. Since $\pi \circ \alpha = \pi$, we conclude that $\pi'(y) \circ \alpha'(x) = \pi'(x)$; that is, $\pi'(y)^{-1} \circ \pi(x) = \alpha'(x)$, which is a positive linear isomorphism. It follows from Proposition $8.2$ that $M/G$ is orientable.
Conversely, if $M/G$ is orientable , we take arbitrarily $\alpha \in G$ and $x \in M$. Let $y = \alpha(x)$. Then $\pi(x) = \alpha(y)$. By proposition $8.2$, the isomorphism $\pi'(y)^{-1} \circ \pi'(x)$ is positive. But this isomorphism coincides with $\alpha'(x)$. It follows that $\alpha$ is positive, which completes the proof.

Why $\pi(x) = \alpha(y)$? I think this doesn't make sense because $\pi(x) \in M/G$ and $\alpha(y) \in M$ (recall that $\alpha \in G$, i.e., $\alpha$ is a diffeomorphism of class $\mathcal{C}^k$ in $M$).

I think you are correct. What is meant is that $\pi(x)=\pi(y)$. Therefore you can compute $\pi^\prime(y)^{-1}\circ\pi^\prime(x)$ because $\pi^\prime(y)$ and $\pi^\prime(x)$ both have value in $T_{\pi(x)}M/G$. Finally as it is said $\pi^\prime(y)^{-1}\circ\pi^\prime(x)=\alpha^{\prime}(x)$ which is positive so you can conclude.
