A very specific example: for G=Z and n=m=1, it is the map S^1 \times S^1 \to CP^\infty which maps S^1 \times S^1 to the 2-skeleton S^2 of CP^\infty by a degree 1 map.

Here's a general but less geometric way of understanding it. K(G,n) can be constructed as the realization of the simplicial abelian group associated to the chain complex with Z in degree n and 0 elsewhere under the Dold-Kan correspondence. By a form of Eilenberg-Zilber, the levelwise tensor product of two simplicial abelian groups is, up to natural weak equivalence, the same as the ordinary tensor product of chain complexes. But the ordinary tensor product takes K(G,n) and K(G,m) to K(G \otimes G,m+n), and there is a natural map from K(G,n) \times K(G,m) to the levelwise tensor product given levelwise by the natural map from a product to a tensor product. Composing all of this together with a ring multiplication map G \otimes G \to G giving K(G \otimes G,m+n)\to K(G,m+n) should give the cup product. With some unraveling that I don't have time to do right now but I invite someone else to figure out, you should be able to turn this into a reasonably explicit map on the simplicial abelian group level coming from the Alexander-Whitney map.

A very specific example: for $G=\mathbb{Z}$ and $n=m=1$, it is the map $S^1 \times S^1 \to CP^\infty$ which maps $S^1 \times S^1$ to the 2-skeleton $S^2$ of $CP^\infty$ by a degree 1 map.

Here's a general but less geometric way of understanding it. $K(G,n)$ can be constructed as the realization of the simplicial abelian group associated to the chain complex with $\mathbb{Z}$ in degree n and 0 elsewhere under the Dold-Kan correspondence. By a form of Eilenberg-Zilber, the levelwise tensor product of two simplicial abelian groups is, up to natural weak equivalence, the same as the ordinary tensor product of chain complexes. But the ordinary tensor product takes $K(G,n)$ and $K(G,m)$ to $K(G \otimes G,m+n)$, and there is a natural map from $K(G,n) \times K(G,m)$ to the levelwise tensor product given levelwise by the natural map from a product to a tensor product. Composing all of this together with a ring multiplication map $G \otimes G \to G$ giving $K(G \otimes G,m+n)\to K(G,m+n)$ should give the cup product. With some unraveling that I don't have time to do right now but I invite someone else to figure out, you should be able to turn this into a reasonably explicit map on the simplicial abelian group level coming from the Alexander-Whitney map.

A very specific example: for G=Z and n=m=1, it is the map S^1 \times S^1 \to CP^\infty which maps S^1 \times S^1 to the 2-skeleton S^2 of CP^\infty by a degree 1 map.

A very specific example: for G=Z and n=m=1, it is the map S^1 \times S^1 \to CP^\infty which maps S^1 \times S^1 to the 2-skeleton S^2 of CP^\infty by a degree 1 map.

Here's a general but less geometric way of understanding it. K(G,n) can be constructed as the realization of the simplicial abelian group associated to the chain complex with Z in degree n and 0 elsewhere under the Dold-Kan correspondence. By a form of Eilenberg-Zilber, the levelwise tensor product of two simplicial abelian groups is, up to natural weak equivalence, the same as the ordinary tensor product of chain complexes. But the ordinary tensor product takes K(G,n) and K(G,m) to K(G \otimes G,m+n), and there is a natural map from K(G,n) \times K(G,m) to the levelwise tensor product given levelwise by the natural map from a product to a tensor product. Composing all of this together with a ring multiplication map G \otimes G \to G giving K(G \otimes G,m+n)\to K(G,m+n) should give the cup product. With some unraveling that I don't have time to do right now but I invite someone else to figure out, you should be able to turn this into a reasonably explicit map on the simplicial abelian group level coming from the Alexander-Whitney map.