The mod $2$ cohomology of $\mathrm{SO}_n$ is not an exterior algebra as soon
as $n\geq3$ (there is a unique non-zero element in degree $1$ whose square is
non-zero, think of the case $\mathrm{SO}_{3}$ which is $3$-dimensional real
projective space) while the homology ring is an exterior algebra.
Another example is $K(\mathbb Z,2)$ (aka $\mathbb C\mathbb P^\infty$), its
integral cohomology ring is a polynomial ring on a degree $2$-generator while
its homology ring is the free divided power algebra on a degree $2$-generator.
Poincare duality is very clearly treated, with real coefficients, via de Rham cohomology, in Spivak's Differential geometry vol. 1, the idea of using open covers with contractible intersections, anticipating sheaf theory,, apparently being due to Weil; and also in Bott - Tu's Differential forms in algebraic topology. Both are recommended.
It is also treated over arbitrary coefficient domains in the appendix to Milnor and Stasheff's Characteristic classes. Anything by Milnor is recommended.
If you just visualize a polyhedron, and its first barycentric subdivision, i.e. placing a new vertex in the center of every face, and forming a new face from the union of all new subtriangles adjacent to a given vertex, you may see the duality arising from a triangulation of a manifold. Thus the theorem is a "obvious" generalization of the duality of the Platonic solids.
The simplest argument I ever heard was in a conversation between John Morgan and Simon Donaldson, at Bob Friedman's house. John said he had a simple proof of Poincare duality using Morse functions, and Simon replied, while turning his hands over, "of course you just turn the Morse function upside down".
If you read the description of the homotopy type of a CW complex in terms of the critical points of Morse functions, say in Milnor's notes on Morse theory, you will learn that a single d cell is added each time we pass a critical point of index d. Since turning a function upside down changes a critical point of index d into one of index n-d, where n is the dimension, we are "done", (modulo the non trivial question of tracing the boundary operators, as noted by a comment below).
I have not read Milnor's notes on the h cobordism theorem, so I do not know if this is the same proof given there, but it is a book on applications of Morse theory. I would suggest the moral is that a young person could do worse than to learn Morse theory.
Best Answer
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.