at.algebraic topology - Understanding the product in topological K-theory

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 $mathbb{Z}left[Hright]/(H-1)^2to K(S^2)$ is an isomorphism of rings. So $tilde{K}(S^2)cong mathbb{Z}left[H-1right]/(H-1)^2$, 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
$K(X)otimes K(X)to K(Xtimes X)to K(X)$.