I'm trying to understand precisely why and how some closed curves are not boundaries while their "doubles" are. I know that taking any curve connecting antipodal points on $S^2$ projects to a closed curve $\gamma$ in $\mathbb R \mathbb P^2$ that generates $\pi_1 (\mathbb R \mathbb P^2)$. The peculiar claim is that although $\gamma$ is not a boundary of any region of $\mathbb R \mathbb P^2$, $\gamma ^2$ is. I've read a bunch of questions and answers on MSE and watched a bunch of GIFs, and I can now visualise pretty well why $\gamma ^2$ is a boundary. (The belt trick worked pretty well for me.) I'm still pretty disoriented (hah) about why $\gamma$ is necessarily not a boundary. The closest I verbal explanation I have encountered is an excerpt from Gray's Worlds out of Nothing, which you see below. I'd much like a surgical explanation of both why $\gamma$ does not disconnect $\mathbb R \mathbb P^2$, and (less important but also much desired) exactly how $\gamma ^2$ does disconnect $\mathbb R \mathbb P^2$.
As was recommended in several MSE answers, I read page 170 in Stillwell's Classical Topology and Combinatorial Group Theory. I'll add more places I tried I need be.
I'd like to stress that I spent several hours reading drawing and thinking before posting this, and I'm aware of the relevant MSE questions, but I want to understand exactly what's going on here and so far I'm unable. Please do not reply tersely or vaguely :)
You can think to $\mathbb{RP}^2$ as the Moebius strip with a disk glued along the boundary: if you draw $\mathbb{RP}^2$ as the square with the edges identified and you take out a neghborhood of an edge (that is a disk) you can see it.
From this viewpoint it is clear the middle circle of the Moebius strip does not disconnet it (and thus it doesn't disconnect $\mathbb{RP}^2$).
While the boundary of the Moebius strip is exactly the double of the middle circle and it divide $\mathbb{RP}^2$ in a disk and in the Moebius strip.