These are indeed not the same in most cases (including when $X=S^1\vee S^1$ or $X=S^1\times S^1$) for the rather trivial reason that the first action is defined using $\bar{\gamma}$ and the second is defined using $\gamma$ (so at least when acting on the basepoint $\tilde{x}_0$, the action of $[\gamma]$ by the first definition corresponds to the action of $[\gamma]^{-1}$ by the second). I'm guessing that what Hatcher had in mind, though, is the version of the first action which is defined using a lift of $\gamma$, not a lift of $\bar{\gamma}$. Of course, that makes the first action a right action rather than a left action, so the actions cannot possibly be the same when $\pi_1(X,x_0)$ is nonabelian (given that the actions are faithful). But for an abelian group, right and left actions are the same, and so this still leaves a nontrivial question of whether they coincide when $\pi_1(X,x_0)$ is abelian.
Let $X,Y$ be nonempty path-connected spaces, $\alpha \in H^p(X;k), \beta \in H^q(Y;k)$ for some field $k$.
Then you get cohomology classes $\alpha \otimes 1 \in H^p(X;k)\otimes H^0(Y;k) \subset H^p(X\times Y;k)$ and $1\otimes \beta \in H^0(X;k)\otimes H^q(Y;k) \subset H^q(X\times Y;k)$.
I claim that the inclusion $i:X\to X\times Y$ at any basepoint of $Y$ sends $\alpha\otimes 1$ to $\alpha$ and $1\otimes \beta$ to $0$. If you apply this in your particular situation with $X=Y= \mathbb RP^{n-1}$ , $\alpha_1$ and $\alpha_2$, you get the desired result.
To prove the claim, you can use the projection $p:X\times Y \to X$ and notice that $p\circ i = id_X$, so that when you look at cohomology, $i^*\circ p^* = id$ on cohomology. In particlar, $i^*(\alpha\otimes 1) = i^*(p^*(\alpha)) = \alpha$. Here, I used $p^*\alpha = \alpha\otimes 1$, but this is essentially part of the Künneth theorem.
Similarly, we may use the projection $q: X\times Y\to Y$ and notice that $q\circ i$ is a constant map, hence it induces $0$ on cohomology. It follows that $0= i^*(q^*(\beta)) = i^*(1\otimes \beta)$; using $q^*\beta = 1\otimes \beta$, which again is essentially part of the Künneth theorem.
This solces the question of "why does the inclusion act like that on $\alpha_1, \alpha_2$ ?"
Then for the second question (how the next sentence follows), consider then $i^*h^*\alpha = i^*(k_1\alpha_1+k_2\alpha_2 ) = k_1i^*\alpha_1 + k_2i^*\alpha_2 = k_1\alpha$ by the previous computation, and $h\circ i$ is a homeomorphism (by what has been said before).
Therefore $i^*h^* = (h\circ i)^*$ is an isomorphism on cohomology, and since $\alpha\neq 0$, it must be that $i^*h^*\alpha \neq 0$, i.e. $k_1\neq 0$.
Best Answer
Fix the two global orientations and call them $+1$ and $-1$. This gives you a map $\tilde M \to \{+1, -1\}$ which is continuous and surjective. So $\tilde M$ is disconnected. Conversely, the inverse images of $+1$ and $-1$ are each homeomorphic to $M$ so those are connected.
To see why the map is continuous you use the definition of the topology to show that $\tilde M$ is orientable (in particular it is locally orientable). So every point has a neighbourhood with a compatible orientation. Meaning $+1$ points have only $+1$ points in a neighbourhood of them and vice versa.