One can notice that your hypothesis is equivalent to say that the closed fiber of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ has dimension $0$ because this closed fibers is equal to $V(\varphi(\mathfrak m)S)$, and $V(I)$ of an ideal $I$ in $S$ is reduced to the closed point if and only the nilradical of $I$ is equal to $\mathfrak n$ which is equivalent to $I$ being $\mathfrak n$-primary as $\mathfrak n$ is maximal.

So one could call your homomorphism ''of relative dimension $0$'', except that I am not sure whether the same condition holds over the other prime ideals of $R$.

Don't you need some noetherian condition to get the finiteness of the length over $S$ ?

**Update** The dimension condition on the closed fiber doesn't propagate to other fibers. Let us consider the following example. Let $k$ be a field, $R=k[[x,y]]$, $S=k(z)[[x,y]]$ et let $\phi$ be the inclusion $R\subset S$. Then $\mathfrak mS=\mathfrak n$. Now I claim that the generic fiber of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ has positive dimension. Otherwise, $S\otimes_R \mathrm{Frac}(R)$ would be a field. But this tensor product is the localization of $k(z)[[x,y]]$ with respect to the multiplicative subset $k[[x,y]]\setminus \lbrace 0 \rbrace$. It is easy to see that there is not enough denominators to make the localization into a field (e.g. $x+zy$ is not invertible in the localization).

So, one can only say "of relative dimension $0$ at the closed fiber". And yes this sounds too long.