I was considering the complex K-theory of $T^2$ and I found, using the split exact sequences associated with the pair $(T^2,S^1 \vee S^1)$ that
$\widetilde{K}^0(T^2)\cong\mathbb{Z}$ and $\widetilde{K}^{-1}(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}$,
so that one arrives at
$K^*(T^2)\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus \mathbb{Z}$.
Now my question is about the ring structure here. We know that in $K(T^2)$ we have the same ring structure as in $K(S^2)\cong \mathbb{Z}[H]/(H-1)^2$, where $H$ is the standard degree one line bundle over $S^2$, and hence we understand how to multiply some factors since the product in $K^*$ reduces to that in $K$ when multiplying degree zero elements, but I can't seem to understand how it works in general. I would appreciate any pointers on how to derive the ring structure in this case.
Thanks in advance!
You have to take more information from the exact sequence. Indeed it tells you what $K^*(T^2)$ is but also where its elements "come from".
For instance, the sequence tells you that $\tilde K^0(S^2)\to \tilde K^0(T^2)$ induced by the quotient map $p:T^2\to S^2$ is an isomorphism.
Similarly, by Bott periodicity, this holds in all even degrees.
In odd degrees, you know that the restriction map $K^{2i+1}(T^2)\to K^{2i+1}(S^1\vee S^1)$, induced by the inclusion $i:S^1\vee S^1\to T^2$, is an isomorphism.
But both $K^*(S^2)\to K^*(T^2)$ and $K^*(T^2)\to K^*(S^1\vee S^1)$ are also ring maps.
So for instance let's take an odd degree element $x\in K^{2i+1}(T^2)$ and an even degree one, $y\in K^{2j}(T^2)$. Then $y= p^*(z)$ for some $z\in K^{2j}(S^2)$, therefore $i^*(xy) = i^*(x) i^*(p^*(z))$. Now $p\circ i$ is the constant map, so $i^*p^*$ is exactly the map induced by $S^1\vee S^1\to *\to S^2$, so if $z \in \mathbb Z[H]/(H-1)^2$, $i^*p^*(z)$ corresponds to taking the degree $0$ monomial of $z$, so it's just $i^*(x) \epsilon(z) = i^*(\epsilon(z)x)$ because $\epsilon(z)\in\mathbb Z$
But $i^*$ is an isomorphism in odd degrees, so $xy = \epsilon(z)x$. So you're fixing the isomorphism $K^{2j}(T^2)\cong \mathbb Z^2$, you forget one of the coordinates of $y$ and you multiply $x$ by that (which is an integer)
The rest of the ring structure is determined similarly : if you multiply two elements of even degree, they'll come from elements of even degree in $S^2$ and so this is entirely controlled by $K^*(S^2)$.
The last thing is what happens when you multiply two odd degree elements?
EDIT : as Tyrone pointed out, the argument I had written here does not work to compute the product of odd degree elements.
In fact, as he suggested, there is a Künneth formula, as $K^*(S^1)$ is flat over $K^*$, that gives $K^*(T^2)\cong K^*(S^1)\otimes K^*(S^1)$, as rings and this allows for a good understanding of the ring structure on $K^*(T^2)$, better than what I tried to do here.
For a reference for that Künneth formula, one can check out the references of this nLab page, see e.g. Theorem (Boa95, 4.19) 2.12
So, to summarize :
-multiplication of even degree elements is controlled by $K^*(S^2)$
-multiplication of elements of different parity is "trivial" : if $x$ is odd and $y$ is even, then $xy = \epsilon(y)x$, where $\epsilon$ is the composite $K^*(T^2)\to K^*(\mathrm{pt})\to K^*(T^2)$