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.
- Look at the fundamental polygon for the Klein bottle: $abab^{-1}$.
This is a $\Delta$-complex with one 0-simplex, three 1-simplicies, and two 2-simplicies.
Note that you have more flexibility in gluing when using a CW structure. In particular, you can construct the Klein bottle with one 0-cell, two 1-cells, and one 2-cell (think of just using a sheet of paper and gluing the edges as instructed in the fundamental polygon, so you don't have to worry about the edge "$c$" in particular).
$\pi_1(K) \cong \langle[a],[b] : [a][b] = [b][a]^{-1} \rangle$. This can be done via Van Kampen and decomposing the polygon (of $K$, without the edge $c$) into two open sets $A$ and $B$ where you can take $A$ to be the union of all edges thickened a bit, and $B$ to be the inside of the polygon with a bit of overlap with $A$ so that $A \cap B$ looks like a frame.
Look at the above picture and do the computation. You should get, with coefficients in $\mathbb{Z}$, $H_0(K) = \mathbb{Z}$, $H_1(K) = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, and $H_2(K) = 0$ (and higher groups vanish too). A sanity check is: $K$ is connected, a surface, and nonorientable, forcing $H_0$ to be $\mathbb{Z}$ and $H_2$ to be $0$. We calculated the fundamental group and Hurewicz tells us $H_1$ is the abelianization of $\pi_1$.
The complex looks like (due to the above CW structure)
$$0 \to \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to 0$$
The only nonzero map is the one from $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$. This, from the picture above, is given by (WLOG) multiplication by $2$ in the first coordinate and $0$ in the second, assuming the first $\mathbb{Z}$ is generated by $a$ and the second by $b$. Now, that you have all the maps, you will get the same answer that you got in $(3)$.
See Hatcher (for example Theorem 2.35).
See Hurewicz theorem or Hatcher 2.A. The basic intuition you should go in with is that $f: I \to X$ can be viewed as both a path and a singular 1-simplex.
Best Answer
The issue is that a quotient of free abelian groups need not be free. Take, for instance, the quotient of $\mathbb{Z}$ by the subgroup $2\mathbb{Z}$. The quotient is $\mathbb{Z}_2$, and this has torsion.