Assume $M$ is a right $R$-module, and $N$ a left $R$-module, for a unital ring $R$. Furthermore, assume that $M'$ is a submodule of $M$.
Can something general be said of when exactly $$m' \otimes n \in M' \otimes_{R} N$$ will agree with $$m' \otimes n \in M \otimes_{R} N$$ for $m' \in M'$?
For example: $$2 \otimes 1 = 0 \in \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2$$ but $$2 \otimes 1 \neq 0 \in 2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2$$ where $$2\mathbb{Z} \triangleleft \mathbb{Z}$$ so that $2\mathbb{Z}$ is a submodule of $\mathbb{Z}$.
My example is a special case, ofc, since one of the arguments is the ring itself in one case, and one of its ideals in the other case, but I believe it illustrates what I am asking.
It's unclear what you mean by "agree with" since those are elements of two different modules. The inclusion $M' \hookrightarrow M$ induces a natural map $M' \otimes N \to M \otimes N$ which sends $m' \otimes n \in M' \otimes N$ to $m' \otimes n \in M \otimes N$, so they always "agree" in that sense.
However, what your example shows is that even though the inclusion $M' \hookrightarrow M$ is an injection, this doesn't imply that the induced map $M' \otimes N \to M \otimes N$ is an injection. This has to do with the failure of $N$ to be flat. In general, one of the several equivalent definitions of flatness is that $N$ is flat iff $(-) \otimes N$ preserves all injections.
Over a Dedekind domain (which includes $\mathbb{Z}$), a module is flat iff it is torsion-free.