The Chern classes give a map $f : BU \to \prod_n K(\mathbb{Z},2n)$, which is a rational equivalence. However, it is not an equivalence over $\mathbb{Z}$ because the cohomology of $BU$ is just a polynomial algebra and has no Steenrod operations. In particular, the generators of the homotopy groups $\pi_{2n}(BU)=\mathbb{Z}$ will not map to generators of the homotopy groups $\pi_{2n}(K(\mathbb{Z},2n))=\mathbb{Z}$. Another way to say this is that the duals of the Chern classes in homology are not in the image of Hurewicz; only certain multiples of them are. What multiples you have to take is determined by the order of the k-invariants of $BU$, which are certain Steenrod operations of the fundamental classes of $K(\mathbb{Z},2n)$.

Steenrod operations can be understood as obstructions to the cup product on ordinary cohomology being strictly commutative. On the other hand, the fact that $f$ is not an equivalence can also be understood as an obstruction to addition in K-theory being strictly commutative. Indeed, any space with a strictly commutative group structure is a product of $K(\pi_n,n)$'s under the map given by a right inverse of the Hurewicz map.

So in some sense you could say that products in cohomology are only homotopy-commutative and sums in K-theory are only homotopy-commutative "for the same reason". Is there some deeper story behind this? I don't know exactly what I'm asking for, but I'd like to get a better understanding of what's going on in this picture.