Differential geometric interpretation of cohomology I'm not sure whether this is an appropriate question for this forum. I'm afraid that this is not a research level question however:
1. It's about reference request therefore the answer does not require any tedious computation.
2. As it is reference request, I think that it is better to ask it here instead of mathstackexchange since here there are more experts familiar with good literature.
3. I believe that this discussion may be useful not only for me.
So what I would like to ask is some good reference for differential geometric interpretation of cohomological stuff like for example: representatives for characteristic classes in geometric terms, cross products, cap product, cup product for deRham forms, pairing with the fundamental class of $M$ as integration over $M$, and Poincare duality. 
 A: For geometric interpretations of cup product and Poincaré duality let us assume that in the following (dual) homology classes are representable by smooth submanifolds (and everything is orientable). Then we can look at intersection theory.
For an integral class $x\in H^i(M;\mathbb R)$ we have that the Poincaré dual $x^* \in H_{m-i}(M,\partial M;\mathbb R)$ is represented by any submanifold which gives, by counting intersections with $i$-manifolds, the same homomorphism on $H_i(M;\mathbb R)$ as $x$. This is what allows you to understand cohomology classes geometrically.
This geometric situation behaves natural in some situations, e.g. consider the inclusion of a submanifold $i:N\to M$. Then we can pull back homology classes in the above sense, but also with the shriek map:
$$ H_i(M ,\partial M)\to H^{m-i}M \stackrel {i^*}\to H^{m-i}N \to H_{n-m+i}(N,\partial N) \to H_{n-m+i}(N,\partial N)/tors.$$
Claim: This map coincides with the above sense, i.e. the map 
$$[(X,\partial X)\hookrightarrow (M,\partial M)] \mapsto [(X\pitchfork N,X\pitchfork \partial N) \hookrightarrow(N,\partial N)],$$ after representing (if possible) $x\in H_i(M,\partial M)$ by an $i$-submanifold $X\hookrightarrow M$ which is transverse to $(N,\partial N)$.
Proof: In terms of dimensions and transverse intersections this makes sense. Note that the codomain is canonically isomorphic to $Hom(H_{m-i}N,\mathbb Z)$, as the map $H_{n-m+i}(N,\partial N) \to H^{m-i}N \to Hom(H_{m-i}N,\mathbb Z)$ factors. In other words the image of $x$ in $H_{n-m+i}(N,\partial N)$ is determined by what it does as a homomorphism on $H_{m-i}N$ up to torsion. To check that the above map does precisely the same as the pullback, defined on the $Hom$ quotient, we just need the above intersection counting. Let $Y^{m-i}\subset \mathring N$ be a submanifold transverse to $(X\cap N)$, then $Y\hookrightarrow N \hookrightarrow M$ is transverse to $X$, and everything evaluates already on $N$.
Okay, now that we got familiar with Poincaré duality we can now look at a Cup product interpretation on homology (which will actually just be a generalization of the above). As above we want $x \in H^iM, y\in H^jM$ and $x^*,y^*$ denote their duals, represented by transverse submanifolds $X,Y$. Then we have:
$$ x \smile y = [X]^* \smile [Y]^* = [X \pitchfork Y]^*,$$
i.e. Theorem: Cup product is dual to transverse intersection.
You should have a good idea in your head now about this interpretation. You should do some own calculation to get more familiar (or excited) with this interpretation.
To combine this with characteristic classes, you will get some insights by noting that the Thom class of the normal bundle of $N \hookrightarrow M$ in $M$ is Poincaré dual to the class represented by $N$. By looking at $M$ the total space of a vector bundle over $N$ (or rather the disk bundle for compactness), you see e.g. why the Euler class of the tangent bundle evaluates to the self intersection of the manifold. This also allows further playing around and applying to other cases.
For references see e.g. Topology books of Bott & Tu, Guillemin & Pollack, Bredon, stratifold theory (Kreck), and you should definitely check out Hutchings' nice notes on this.
I really hope this monologue also helps you and not only my desire to procrastinate.
