I think people should ignore all this nonsense about counting holes and just look at what actually happens in more examples.

In particular, your intuition about top homology having something to do with enclosing volumes is not quite correct. I interpret this to mean that you have in mind a manifold which is the boundary of another manifold (the same way that the sphere $S^n$ is the boundary of the disk $D^{n+1}$), and it's not true that a manifold has to be a boundary in order to have nonvanishing top homology. The simplest closed counterexample is $4$-dimensional: the complex projective plane $\mathbb{CP}^2$ is a $4$-manifold with $H_4 = \mathbb{Z}$, but it's known not to be the boundary of a $5$-manifold.

Whether top homology vanishes or not instead has to do with orientability.

It's possible to directly visualize the case of $\mathbb{RP}^2$, so let's do it. In this case $\pi_1 \cong H_1 \cong \mathbb{Z}_2$, so the goal is to visualize why there's some loop which isn't trivial but such that twice that loop is trivial. Visualize $\mathbb{RP}^2$ as a disk $D^2$, but where antipodal points on the boundary have been identified. We'll consider loops starting and ending at the origin.

I claim that a representative of a generator of $\pi_1 \cong H_1$ is given by the loop which starts at the origin, goes up to the boundary, gets identified with the opposite point, and goes up back to the origin. Try nudging this loop around for a bit so you believe that it's not nullhomotopic: the point is that you can't nudge it away from the boundary because the two (antipodal, hence identified) points it's intersecting the boundary at can never annihilate.

Now we want to visualize why twice this loop is nullhomotopic. It will be convenient to nudge the loop so that it hits the boundary at four points, which come in two antipodal pairs $A, A', B, B'$, so that the loop hits them in that order before returning to the origin. At this point it would be helpful to draw a diagram if you haven't already; the disk, and the two loops in it, should look a bit like a tennis ball in profile. Now, nudge the loop so that $A, B'$ get closer together, and hence, since they're constrained to be antipodes, $A', B$ also get closer together. Eventually you'll have nudged them enough that you'll see that you can finally pull the curve away from the boundary: as you do so, $A', B$ annihilate each other, and then $A, B'$ annihilate each other.

A similar visualization works for the Klein bottle, visualized as a square with its sides identified appropriately.