Let $M$ be a $G$-module. Then there's an obvious multiplication map $${\mathbb Z}G\otimes_{{\mathbb Z}H}M\rightarrow M$$

Taking homology gives $$H_*(G,{\mathbb Z}G\otimes_{{\mathbb Z}H}M)\rightarrow H_*(G,M)$$

To compute the group on the left, start with a projective resolution $P$ of ${\mathbb Z}$ over ${\mathbb Z}G$, then take the homology of
$$P\otimes_{{\mathbb Z}G}{\mathbb Z}G{\otimes_{\mathbb ZH}}M=P\otimes_{{\mathbb Z}H}M$$

Notice that $P$ is also a projective resolution of ${\mathbb Z}$ over ${\mathbb Z}H$, so the homology you've just computed is in fact equal to $H_*(H,M)$.

Thus the map in 2) is actually a map $H(H,M) \rightarrow H(G,M)$ (where I have supressed the star subscript so that the MathJax will render correctly) and is in fact the transfer map.

To make the same argument work for a ring inclusion $A\rightarrow B$ you'd need to know that a projective $B$-resolution is also a projective $A$-resolution. This works, for example, if $B$ is $A$-free.

Let $M$ be a $G$-module. Then there's an obvious injection $$M\rightarrow {\mathbb Z}G\otimes_{{\mathbb Z}H}M$$

Taking homology gives $$H_\bullet(G,M)\rightarrow H_\bullet(G,{\mathbb Z}G\otimes_{{\mathbb Z}H}M)$$

To compute the group on the right, start with a projective resolution $P$ of ${\mathbb Z}$ over ${\mathbb Z}G$, then take the homology of
$$P\otimes_{{\mathbb Z}G}{\mathbb Z}G{\otimes_{\mathbb ZH}}M=P\otimes_{{\mathbb Z}H}M$$

Notice that $P$ is also a projective resolution of ${\mathbb Z}$ over ${\mathbb Z}H$, so the homology you've just computed is in fact equal to $H_\bullet(H,M)$.

Thus the map in 2) is actually a map $H_\bullet(G,M) \rightarrow H_\bullet(H,M)$ and is in fact the transfer map.

To make the same argument work for a ring inclusion $A\rightarrow B$ you'd need to know that a projective $B$-resolution is also a projective $A$-resolution. This works, for example, if $B$ is $A$-free.