Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism.
Looking at just the group structure, it is enough to have that $G\subset \ker f$ to ensure the existence of a group morphism $g$ such that $f=g\circ \pi$.
When is this $g$ in fact a morphism of varieties?
This is Theorem 4 on page 73 of Mumford's Abelian Varieties, where you will also find a proof. Here's the statement: Let $X$ be an abeian variety. There is a 1-1 correspondence between the two sets of objects:
(a) finite subgroups $K\subset X$
(b) separable isogenies $f:X\to Y$, where two isogenies $f_1:X\to Y_1$ and $f_2:X\to Y_2$ are considered equal if there is an isomorphism $h:Y_1\to Y_2$ such that $f_2=h\circ f_1$, which is set up by $K=\text{ker}(f)$, and $Y=X/K$.
This should always be the case. Write $f$ as a composition $f^{\mathrm{insep}} \circ f^{\mathrm{sep}}$, where $f^{\mathrm{insep}}$ is purely inseparable and $f^{\mathrm{sep}}$ is separable. Note that as subgroups, the kernel of $f$ equals the kernel of $f^{\mathrm{sep}}$. Now by the typical construction of a quotient abelian variety, $\pi$ is separable, so $f^{\mathrm{sep}} = h \circ \pi$ for some (separable) isogeny $h$ (this is a standard fact which I think can be proven by looking at towers of extensions of function fields). Then we can take $g = f^{\mathrm{insep}} \circ h$ to get $f = g \circ \pi$. Notice that $g$ and $h$ are the exact same map when considered only as group homomorphisms, which is the map you specified (as group homomorphisms, $f = g \circ \pi = h \circ \pi$). However, over finite fields there are many distinct morphisms of varieties which as maps on sets are the same but which differ by a purely inseparable morphism.
