I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate…
The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_0)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.
If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\left[H\right]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\left[H-1\right]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.
The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?
Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map
$K(X)\otimes K(X)\to K(X\times X)\to K(X)$.
Best Answer
Reduced $K$-groups are ideals of the standard $K$-groups. $\tilde K(X) \subset K(X)$ is the ideal of virtual-dimension-zero elements.
In particular, the reduced K-theory $\tilde K(S^2)$ is not $\mathbb{Z}[H]/(H-1)^2$, but rather the ideal of this generated by $(H-1)$. In particular, any element in this group does square to zero.
Additionally, the "exterior product" isomorphism $f: \tilde K(X) \to \tilde K(X \wedge S^2)$, which is an isomorphism, is not a ring map: it takes an element $x$ to the exterior product $x \wedge (H-1)$. Instead, it satisfies $f(x) f(y) = (H-1) f(xy) = 0$. This is because the suspension is covered by two contractible open subsets, and so all products must vanish.