Torsion elements seem intuitively significant as well as torsion-free elements. But let us start from first principles, finding the conditions under which a chain $c = \sum_{i}a_{i} \sigma_{i} \in C_{n}$ might, in an intuitive sense, be considered equivalent to an n-dimensional hole, and relating these conditions to homology.
Roughly, $c$ is an arrangement of simplices, singular simplices, cells, etc., each of which $\sigma_{i}$ appears $|a_{i}|$ times in a "forwards" orientation if $a_{i}$ is positive and a "reverse" orientation if $a_{i}$ is negative. If this arrangement $c$ is a candidate for being equivalent to a hole, it must be a cycle: that is, an $n$-dimensional chain with no $(n-1)$-dimensional beginning or end. For example, an element of the group of one-dimensional cycles $Z_{1}$ might be a "cyclical" chain of edges with no endpoints, while an element of $Z_{2}$ might be a chain of faces that similarly "goes all the way around" with no outer edge. Now, to evaluate the boundary operator $\partial_{n}:C_{n}\rightarrow C_{n-1}$ at $c$ is essentially to send each $\sigma_{i}$ to the "ending" parts of its $(n-1)$-dimensional boundary, while subtracting the "beginning" parts. Some terms of $\partial_{n}(c)$ might cancel out, as an end of one or more of the $\sigma_{i}$ might be the beginning of another, but for there to be no beginning or end, the terms of $\partial_{n}(c)$ must ultimately sum to zero. Thus the group of cycles $Z_{n}$ is intuitively $\ker(\partial_{n})$.
To show the intuition, then, we should show that quotienting out by $B_{n}=\text{im}(\partial_{n+1})$ gives the particular cycles that are equivalent to "holes" as generators (i.e. members of some choice of generating set) of the resulting homology group $H_{n}=Z_{n}/B_{n}$. While $c \in H_{n}$ is equivalent to the simplest type of hole if it is a torsion-free generator, such as a tunnel shape (like a meridian of a torus) if $n=1$, or a cavity shape (like one sphere in a wedge sum of spheres) if $n=2$, we can also provide some sense of what generators of finite order represent geometrically.
Clearly $c$ is not equivalent to a hole if it is filled by a chain ${c}'$ of dimension $n+1$: that is, if $c = \partial_{n+1}({c}')$, so $c\in B_{n}$ and $c$ is trivial in $H_{n}$. Conversely, $c$ is equivalent to some kind of hole if it alone is not filled by any $(n+1)$-dimensional chain. The simplest kind of hole mentioned above can be considered a cycle no nonzero multiple of which is filled by such a chain. If $c$ is equivalent to one of these holes, then it is a generator of $Z_{n}$ no nonzero multiple of which is in $B_{n}$, and therefore a torsion-free generator of $H_{n}$. Another kind of hole is a cycle that is not filled by a $(n+1)$-dimensional chain, but some multiple of which is so filled; the nontrivial element of $H_{1}(\mathbb{R}\text{P}^{2})$ is this kind. When $c$ is a generator of $Z_{n}$ such that some $k\geq2$ is the lowest positive integer with $kc$ a boundary, and thus a generator of order $k$ of $H_{n}$, it satisfies the intuition for this kind of hole, which we call an order-$k$ hole. The first kind of hole represented by an element of infinite order, following this terminology, is referred to as an order-$\infty$ hole, and an order-$1$ hole is represented by a cycle that does not determine a hole at all.
Recognize that not all elements of $H_{n}$ belong to our chosen generating set, so they are sums of holes just as $C_{n}$ consists of sums of cells or simplices. Also recognize that many cycles could determine the same hole (or sum of holes); take two meridians of a torus, for example. This relationship corresponds with cycles being two different representatives of the same homology class, and thus having their difference (in $Z_{n}$) belonging to $B_{n}$, so their difference is a non-hole and thus negligible from the standpoint of holes.
This notion of order-$k$ holes allows us to more intuitively extend the significance of torsion-free generators to torsion generators under the classical homological conceptualization of a hole as "a cycle that is not a boundary." For $k\geq2$, an order-$k$ hole is "a cycle that is not a boundary but is $\frac{1}{k}$ a boundary." It is $\frac{1}{k}$ filled, so $k$ of it together would be filled. An order-$\infty$ hole is, by extension, "a cycle that is neither a boundary nor a fraction of a boundary". In summary, homology groups count holes and the fractions by which they are filled.
Now on the issue of $0$-dimensional homology, we proceed by analogy. $H_{2}$ measures cavities, or $2$-cycles unfilled by $3$-chains. $H_{1}$ measures tunnels, or $1$-cycles unfilled by $2$-chains. So $H_{0}$ measures gaps, or $0$-cycles unfilled by $1$-chains. One would expect such gaps to be pairs consisting of one beginning point and one endpoint, with the region between them unfilled. The number of gaps, and thus the rank of $H_{0}$, would most intuitively be the number of path-components minus one, as is clear if one "lines up" the path components. But if we take the obvious route and take unreduced homology, we get a torsion-free element of $H_{0}(X)$ even for $X$ path-connected; the reason for this is that all $0$-chains are cycles, including single points that end up representing nontrivial homology classes. Our above intuition suggests that only chains generated by endpoints minus beginning points, or by chains of the form $\sigma-\tau$ with $\sigma$ and $\tau$ individual points, should be cycles, which is true in reduced homology under the new definition of $\partial_{0}$. And the overall effect of replacing $H_{0}$ with $\tilde{H}_{0}$ is felicitous, quotienting out by the "guaranteed" extra torsion-free element to produce a group of desired rank.
EDIT July 18 2015: This answer cites a book by John Stillwell providing an explanation for the torsion of homology groups similar to the one I have given.
Best Answer
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.