The induced map $\overline{\pi}: \operatorname{Hom}(Z_2,Z) \to\operatorname{Hom}(Z_2,Z_2)$?

You have $\varphi=\pi$ going from $A=\mathbb Z$ to $C=\mathbb Z/2\mathbb Z$, hence the induced map $\overline\varphi=\overline\pi$ goes from

$$ \operatorname{Hom}(C,R)=\operatorname{Hom}(\mathbb Z/2\mathbb Z,R)\quad\text{to}\quad\operatorname{Hom}(A,R)=\operatorname{Hom}(\mathbb Z,R) $$

for some module $R$. As you can see, we cannot choose $R$ such that we get the setup from Hungerford, so this is most likely a typo. As you have observed a counterexamples arises regardless if we take $R=\mathbb Z$.

This is not a typo: you are referencing the wrong theorem. You should instead reference to Theorem 4.4, which is the COvariant version of the result you need. Unfortunately, Hungerford uses overbars for both induced maps $\overline{\phi}$. When $\phi: A \to B$, he writes $\overline{\phi}$ both for $$ \overline{\phi} : \operatorname{Hom}(C,A) \to \operatorname{Hom}(C,B) $$ as well as for $$ \overline{\phi} : \operatorname{Hom}(B,C) \to \operatorname{Hom}(A,C). $$ You need the first version. where $C=\mathbb{Z}_2$ is fixed.

Many authors use different notations for these different induced maps precisely to avoid this confusion. It is common to use $\phi_*$ for the covariant version and $\phi^*$ for the contravariant.