What carries over?

As Peter pointed out, the kernel of a map of free $\mathbb{Z}$-modules though free need not have a complement. Indeed each submodule of a free $\mathbb{Z}$-module is free, but a quotient module need not be, for instance $\mathbb{Z}/2\mathbb{Z}$.

The set $\mathrm{Hom}(F,G)$ for free $\mathbb{Z}$-modules need not be free. If $F$ is free of countably infinite rank and $G=\mathbb{Z}$, then $\mathrm{Hom}(F,G)\cong\prod_{j=1}^\infty\mathbb{Z}$ which remarkably is not free over $\mathbb{Z}$. But $F\otimes G$ is free for free $F$ and $G$.

What carries over?

As Peter pointed out, a submodule of a free $\mathbb{Z}$-module though free need not have a complement. Indeed each submodule of a free $\mathbb{Z}$-module is free, but a quotient module need not be, for instance $\mathbb{Z}/2\mathbb{Z}$. Also a $\mathbb{Z}$-module is free if and oly if it is projective; this entails that a kernel of a map of free modules does have a complement.

The set $\mathrm{Hom}(F,G)$ for free $\mathbb{Z}$-modules need not be free. If $F$ is free of countably infinite rank and $G=\mathbb{Z}$, then $\mathrm{Hom}(F,G)\cong\prod_{j=1}^\infty\mathbb{Z}$ which remarkably is not free over $\mathbb{Z}$. But $F\otimes G$ is free for free $F$ and $G$.