Clarifying maps in algebraic number theory

I'd think of it like this.

There is a natural embedding $\mathbb Z\hookrightarrow\mathcal O_K$, and hence a projection $$\mathbb Z\hookrightarrow\mathcal O_K\to\mathcal O_K/\mathfrak m.$$

The kernel of this map is exactly the elements of $\mathfrak m$ which are also in $\mathbb Z$ - i.e. $\mathbb Z\cap\mathfrak m$. We deduce that there is an injection $$\mathbb Z/\mathbb Z\cap \mathfrak m\hookrightarrow\mathcal O_K/\mathfrak m.$$

In the case where $K$ is an imaginary quadratic number field, these two fields are just the residue fields of $\mathbb Z$ and $\mathcal O_K$ with respect to $\mathbb Z\cap \mathfrak m$ and $\mathfrak m$ respectively. Therefore, the map will be surjective if and only if the two residue fields are equal, which happens if and only if $\mathbb Z\cap\mathfrak m$ ramifies or splits in $K$.

This is just a general fact about modules over commutative rings. Note that $\mathcal{O}_K/\mathfrak{m}$ is a $\Bbb Z/\mathfrak{m}\cap\Bbb Z$ module--the action is just multiply by $n\in\Bbb Z$, and clearly it is trivial on $\mathfrak{m}\cap\Bbb Z$, so you can map the latter in the former by looking at $\Bbb Z/\mathfrak{m}\cap\Bbb Z\cdot 1$, as a cyclic $\Bbb Z/\mathfrak{m}\cap\Bbb Z$ submodule of $\mathcal{O}_K/\mathfrak{m}$.