What is known about the intersection pairing on H^{mid}? When we restrict to the torsion-free part of the cohomology of a manifold, the intersection pairing is nondegenerate.  In dimension 2n, this gives a bilinear form on the free part of Hn (symmetric if n is even, skew-symmetric of n is odd).  Supposedly, this classifies simply-connected 4-manifolds pretty well; more precisely, the map from homeomorphism classes of manifolds is at most 2-to-1, and whenever a bilinear form has two preimages, at least one of them isn't smooth.
Broadly, I'm wondering what else is known about this pairing.  (Of course, if anything I've said so far is wrong, please correct me!)  For example, can we obtain the bilinear form associated to a connected-sum (or a product, or whatever other natural ways we can get manifolds) in some nice way?  It seems like in the connected-sum case, Mayer-Vietoris should tell us something -- I'd imagine the isomorphism it gives (when n>1) between Hn(connected-sum)=Hn(M1)+H(M2) is probably natural, but I don't know if we can tell anything about the cohomology ring structure from it.  As for the product of manifolds, I'm pretty sure that the Kunneth formula gives an isomorphism of algebras, so that should take care of it unless I'm missing something...  Also, is this at all related to the Kirby-Siebemann invariant?  (All I really know is that it's related to smoothness/smoothability, but perhaps, for example, we can somehow read off the possible values of the K-S invariant from the bilinear form?)
 A: I think you are right about Mayer-Vietoris. It gives the isomorphism $H_n(sum)=H_n(M_1)\oplus H_n(M_2)$ and since homology classes can be represented by surfaces that don't hit "gluing disk" we get that the intersection form of the sum $Q=Q_1\oplus Q_2$.
If a manifold admits a smooth (or even PL) structure then KS vanishes. 
In dimension 4:
I think that K-S invariant is independent of the intersection form if it is odd. If the form is odd then there're two homeomorphism types of manifolds (Freedman) that are distinguished by the $KS(M)\in H^4(M,\mathbb Z_2)$. 
If the form is even then KS(M)=(1/8)signature mod 2. 
A: Not an answer to your question, but I just want to stress that, whilst the intersection form does more-or-less classify simply-connected topological 4-manifolds, it is far from the whole story in the smooth category. There are many examples of simply-connected topological 4-manifolds which admit infinitely many different smooth structures. (This is the only dimension in which this occurs.)
These different smooth structures are usually detected via Donaldson or Seiberg-Witten invariants. These invariants also enter in the connected sum discussion. Given a connect-sum of topological 4-manifolds X=M#N, as others have said, the intersection forms add: Q(X)= Q(M)+Q(N). Freedman's work on 4-manifolds tells us that there is essentially a converse to this in the topological category: roughly, if a simply-connected topological 4-manifold X has intersection form which can be written as Q(M)+Q(N) for simply-connected 4-manifolds M,N then in fact there is connected sum decomposition X=M#N. 
On the other hand, one can show that for many smooth 4-manifolds M,N the, say, Seiberg-Witten invariants of M#N must vanish. Meanwhile, there are many X for which we know that the invariants are non-zero (e.g., symplectic 4-manifolds thanks to the work of Taubes). This means that there are many examples of smooth 4-manifolds which can be written topologically as a connected sum - by Freedman's work - but not smoothly because the SW invariant is non-zero.
A: The bilinear form associated to a connected sum should be the direct sum of the two pairings, yes. Intuitively, classes in M do not intersect classes in N, so they are orthogonal with respect to the pairing.
A: Just to partly answer your question. For a simply connected closed $4$-manifold the intersection pairing Q in cohomology determines the ring structure. And even if there is torsion in $H_2(M;\mathbb{Z})$, by the universal coefficient theorem or otherwise $H^2(M;\mathbb{Z})$ is free. By Poincare duality $H^3(M;\mathbb{Z})=0$. Therefore, knowing $Q$ is knowing $H^\ast(M;\mathbb{Z})$. And as all the previous answers pointed out (and you yourself) for a connected sum of $M_1$ and $M_2$ the $H^2$ adds up and the ring structure is determined by $Q_1\oplus Q_2$.
