In Step 2, I don't understand the part "for otherwise, two points would belong to the same orbit and the restriction of $\pi$ to $U_\alpha$ would not be injective". What two points? If $g\ne e$ and $U_\alpha\cap gU_\alpha\ne \emptyset$, then $y\in U_\alpha\cap gU_\alpha$. And we also know $x\in U_\alpha$. Either $x\ne y$ or $x=y$ (and it may happen that $x$ is the only point in the intersection). But in either case, I don't understand why the result follows. In the former case, why must $y$ lie in the same orbit as $x$? In the latter case, if $x$ is the only point in the intersection, where does the second point come from?
Assume that $\pi$ is a covering map.
(1) Each $g \in G$ is a covering transformation because $g$ is a homeomorphism such that $\pi \circ g = \pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g \in G \setminus \{ e \}$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_\alpha) \cap U_\alpha \ne \emptyset$, where $g \ne e$.
So let $y \in g(U_\alpha) \cap U_\alpha$. Let $z \in U_\alpha$ be the unique point such that $g(z) = y$. We have $z \ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $\pi(z) = \pi(y)$. This means that $\pi : U_\alpha \to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $\pi : X \to B$ be a covering map with a path connected $X$. Then the group $\mathcal{C}(X,\pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x \in X$ and do the same as above until you get $z, y \in U_\alpha$. Since $\pi \circ g = \pi$, you get $\pi(y) = \pi g(z) = \pi(z)$ which shows that $\pi : U_\alpha \to V$ is not injective.
Munkres applies this to $\pi : X \to X/G$. At this point he only uses the fact $G \subset \mathcal{C}(X,\pi,X/G)$ (see (1) above). That $G = \mathcal{C}(X,\pi,X/G)$ is shown in Step 3.