Let $e_{A}(\phi, M): = l_A(\mathrm{coker}(\phi) ) - l_A(\ker(\phi))$.
In my book it is stated that if $IM = 0 \implies e_{A}(\phi, M) = e_{A/I}(\phi, M)$ and this seems to be obvious for the author. But for me it is not obvious and I have no idea how to prove it.
If $M$ is an $A$-module and if $I$ is an ideal of $M$, then $M/IM$ is an $A/I$-module. Indeed : recall that a structure of $A$-module is the same as a $\mathbf{Z}$-module (an abelian group) structure plus an action of $A$ on $M$, that is, a morphism of rings $A\to \textrm{Hom}_{\textrm{Ab}.}(M,M)$. (This is very important to realize, even if it is not hard.) Saying that $IM = 0$ means then that the kernel of this morphism contains $I$, and then, the morphism $A\to \textrm{Hom}_{\textrm{Ab}.}(M,M)$ passes to the quotient to give a ring morphism $A/I \to \textrm{Hom}_{\textrm{Ab}.}(M,M)$ which is but an $A/I$-module structure on our $M$, according to this "action" interpretation. (This is a more elegant way to do it. But you could have set $\overline{a} m := am$ and told that it does not depend on $a$ as $IM = 0$.)
But when $IM = 0$, then you see that $M/IM = M$ is an $A/I$-module. Now, do you see then why the length doesn't change ? (It is really because in this case $A$ or $A/I$-module is the same. What have you tried by the way ?)