Can the cohomology ring of the two-fold torus be calculated abstractly? In our lectures, we are given an unusual definition of cohomology and cup products which makes explicit calculations a bit tedious (that is, even more tedious than usual).
For the $n$- and $m$-spheres $S^n$ and $S^m$, I know that there’s an isomorphism of rings
$$H^•(S^n×S^m) \overset \simeq \longrightarrow H^•(S^n) \otimes H^•(S^m).$$
Can I use merely this fact, and other basic facts about cohomology like the Mayer–Vietoris sequence, Poincaré duality and the functoriality of cohomology to actually calculate the cohomology of the two-fold torus? What other  facts would I need?
 A: (It is not clear to me what you mean by "abstractly", I take this to mean "using general principles alone", which is only slightly less ambiguous.)
In any case, I propose the following. Our double torus is the connected sum $T^2\# T^2$ (where as usual $T^2=S^1\times S^1$).


*

*There is a general description of $H^\ast(M\# N)$ (with $M$ and $N$ oriented manifolds) in terms of $H^\ast(M)$ and $H^\ast(N)$, which is described in detail for instance here. I think that this can be regarded as a basic fact about cohomology, in fact much more basic than e.g. Poincaré duality. 

*Using the description in 1, we are left with the task of calculating the cohomology ring $H^\ast(T^2)$. But as you say, Künneth's theorem gives $H^\ast(T^2)=H^\ast(S^1)\otimes H^\ast(S^1)=\bigwedge_{\mathbf{Z}}[x,y]$ with $x$ and $y$ in degree $1$. (The description of the cohomology ring of the spheres is obvious.)
It is possible, using simplicial cohomology, to describe the cohomology ring of the double torus using only the definition. This is done for instance in Hatcher's algebraic topology (page 207, Example 3.7). But I regard the approach above as more satisfactory, and I think it would be misleading to think that using the definition alone is the way to do these kinds of computations in general; in my opinion, if one does a generic computation with (co-)homology, one cannot do a lot using the definition alone.
A: I think Qiaochu Yuan's post (https://qchu.wordpress.com/2013/10/12/the-cohomology-of-the-n-torus/) is quite relevant. In particular the first part gives a derivation of the cohomology (group) of the $n$-torus using Mayer–Vietoris . Of course you need more tools to compute the ring structure, but I think Qiaochu did a quite good job at elaborating it as well. So I feel an answer by me might be simply duplicating some parts of what he did. Of course you can use Poincare duality, but the computation of the homology groups takes some effort if you just use Mayer–Vietoris or long exact sequence.  
