You can indeed! By Yoneda, any natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in H^n(K(Z,n);G) coming from the universal class in H^n(K(Z,n);Z), and this will give your desired map K(Z,n) \to K(G,n). However, there's a nice bigger story behind this specific case besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now K(G,n) can be realized as the simplicial abelian group which corresponds to the chain complex with G in degree n and 0 everywhere else. Maps of simplicial abelian groups (mod homotopy) from K(G,n) to K(H,m) are then the same as maps of chain complexes in the derived category, i.e. Ext^{m-n}(G,H). This is 0 except for m=n and m=n+1, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with m=n+1 (coming from the Ext part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of Ext).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from K(G,n) to K(H,m) which are group homomorphisms with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism up to homotopy, since the group structure on cohomology is just the group structure on K(G,n) taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case n=m.

You can indeed! By Yoneda, any natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in $H^n(K(\mathbb{Z}, n); G)$ coming from the universal class in $H^n(K(\mathbb{Z}, n); \mathbb{Z})$, and this will give your desired map $K(\mathbb{Z}, n) \to K(G, n)$. However, there's a nice bigger story behind this specific case besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now $K(G, n)$ can be realized as the simplicial abelian group which corresponds to the chain complex with $G$ in degree $n$ and $0$ everywhere else. Maps of simplicial abelian groups (mod homotopy) from $K(G, n)$ to $K(H, m)$ are then the same as maps of chain complexes in the derived category, i.e. $\operatorname{Ext}^{m-n}(G, H)$. This is $0$ except for $m=n$ and $m=n+1$, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with $m=n+1$ (coming from the $\operatorname{Ext}$ part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of $\operatorname{Ext}$).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from $K(G, n)$ to $K(H, m)$ which are group homomorphisms with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism up to homotopy, since the group structure on cohomology is just the group structure on $K(G, n)$ taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case $n=m$.

You can indeed! By Yoneda, any natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in H^n(K(Z,n);G) coming from the universal class in H^n(K(Z,n);Z), and this will give your desired map K(Z,n) \to K(G,n). However, there's a nice bigger story behind this specific besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now K(G,n) can be realized as the simplicial abelian group which corresponds to the chain complex with G in degree n and 0 everywhere else. Maps of simplicial abelian groups (mod homotopy) from K(G,n) to K(H,m) are then the same as maps of chain complexes in the derived category, i.e. Ext^{m-n}(G,H). This is 0 except for m=n and m=n+1, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with m=n+1 (coming from the Ext part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of Ext).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from K(G,n) to K(H,m) which are group homomorphisms with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism up to homotopy, since the group structure on cohomology is just the group structure on K(G,n) taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case n=m.

You can indeed! By Yoneda, any natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in H^n(K(Z,n);G) coming from the universal class in H^n(K(Z,n);Z), and this will give your desired map K(Z,n) \to K(G,n). However, there's a nice bigger story behind this specific case besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now K(G,n) can be realized as the simplicial abelian group which corresponds to the chain complex with G in degree n and 0 everywhere else. Maps of simplicial abelian groups (mod homotopy) from K(G,n) to K(H,m) are then the same as maps of chain complexes in the derived category, i.e. Ext^{m-n}(G,H). This is 0 except for m=n and m=n+1, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with m=n+1 (coming from the Ext part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of Ext).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from K(G,n) to K(H,m) which are group homomorphisms with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism up to homotopy, since the group structure on cohomology is just the group structure on K(G,n) taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case n=m.