The Eilenberg-Zilber theorem says that for singular homology there is a natural chain homotopy equivalence:
$$S_*(X)\otimes S_*(Y) \cong S_*(X\times Y)$$
The map in the reverse direction is the Alexander-Whitney map. Therefore we obtain a map
$$S_*(X)\rightarrow S_*(X\times X) \rightarrow S_*(X)\otimes S_*(X)$$
which makes $S_*(X)$ into a coalgebra.
My source (Selick's Introduction to Homotopy Theory) then states that this gives $H_*(X)$ the structure of a coalgebra. However, I think that the Kunneth formula goes the wrong way. The Kunneth formula says that there is a short exact sequence of abelian groups:
$$0\rightarrow H_*(C)\otimes H_*(D) \rightarrow H_*(C \otimes D) \rightarrow \operatorname{Tor}(H_*(C), H_*(D)) \rightarrow 0$$
(the astute will complain about a lack of coefficients. Add them in if that bothers you)
This is split, but not naturally, and when it is split it may not be split as modules over the coefficient ring. To make $H_*(X)$ into a coalgebra we need that splitting map. That requires $H_*(X)$ to be flat (in which case, I believe, it's an isomorphism).
That's quite a strong condition. In particular, it implies that cohomology is dual to homology.
Of course, if one works over a field then everything's fine, but then integral homology is so much more interesting than homology over a field.
In the situation for cohomology, only some of the directions are reversed, which means that the natural map is still from the tensor product of the cohomology groups to the cohomology of the product. Since the diagonal map now gets flipped, this is enough to define the ring structure on $H^*(X)$.
There are deeper reasons, though. Cohomology is a representable functor, and its representing object is a ring object (okay, graded ring object) in the homotopy category. That's the real reason why $H^*(X)$ is a ring (the Kunneth formula has nothing to do with defining this ring structure, by the way). It also means that cohomology operations (aka natural transformations) are, by the Yoneda lemma, much more accessible than the corresponding homology operations (I don't know of any detailed study of such).
Rings and algebras, being varieties of algebras (in the sense of universal or general algebra) are generally much easier to study than coalgebras. Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer. Certainly, I feel that I have a better idea of what a ring looks like than a coalgebra. One thing that makes life easier is that often spectral sequences are spectral sequences of rings, which makes them simpler to deal with - the more structure, the less room there is for things to get out of hand.
Added Later: One interesting thing about the coalgebra structure - when it exists - is that it is genuinely a coalgebra. There's no funny completions of the tensor product required. The comultiplication of a homology element is always a finite sum.
Two particularly good papers that are worth reading are the ones by Boardman, and Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology. Although the focus is on operations of cohomology theories, the build-up is quite detailed and there's a lot about general properties of homology and cohomology theories there.
Added Even Later: One place where the coalgebra structure has been extremely successfully exploited is in the theory of cohomology cooperations. For a reasonable cohomology theory, the cooperations (which are homology groups of the representing spaces) are Hopf rings, which are algebra objects in the category of coalgebras.
Best Answer
There is an interpretation of homology using derived categories, and you can find a brief treatment around remark 3.3.10 in Kashiwara-Schapira "Sheaves on Manifolds". If $M$ is an $n$-manifold and $a: M \to \ast$ is the tautological map to a point, then the homology of $M$ with coefficients in an abelian group $F$ is the homology of the complex $Ra_! a^! F$. If we write $or_M = H^{-n}(a^!F)$ for the orientation sheaf, then the counit of the adjunction induces an "integration" map $Ra_! a^! F \to F$ that induces a map $H^n_c(M,or_M) \to F$. When $F$ is a real vector space and $M$ is smooth, you can relate this to differential forms, because the orientation sheaf is quasi-isomorphic to a suitable version of the de Rham complex. Softness of the de Rham sheaves implies you get an isomorphism $H^n_c(M, or_M) \to \frac{\Gamma_c(M, \Omega^n \otimes or_M)}{d\Gamma_c(M,\Omega^{n-1} \otimes or_M)}$. Stokes's theorem on $M$ implies a comparison between the "integration" map and "honest integration" of forms (or densities), in particular, the fact that $d\Gamma_c(M,\Omega^{n-1} \otimes or_M)$ is annihilated by honest integration.
I don't think Stokes's theorem is a statement whose proof is made easier with derived categories. Instead, I would rather say that it implies the fact that honest integration actually yields a morphism in a derived category.
Here's the setup: If we are given a smooth manifold $M$ and a $k$-form $\omega$, we can obtain a real number from any smooth map $f$ from the standard $k$-simplex $\Delta$ to $M$, by integrating $f^\ast\omega$ along $\Delta$. This produces a map of graded real vector spaces $\int_M: \Omega^\ast(M) \to \operatorname{Hom}(\operatorname{Sing}_\ast(M),\mathbf{R})$.
Here's Stokes's theorem: $\int_M$ is in fact a map of cochain complexes.
If you want to prove the theorem efficiently, you can use naturality of pullback to reduce to a simpler statement about forms on $\Delta$ itself. There will always be a step where you have to use properties of integrals. It is common to invoke Fubini's theorem (e.g., the short proof in Guillemin-Pollack), but you can do a lower-tech proof using the fact that polynomial forms uniformly approximate smooth forms on the simplex.