I am working with modules, but I guess this question is valid with any abelian category.
Let $R$ be a ring, and $F$ a functor. Let $A, M$ be $R$-modules and $f : A \longrightarrow M$ a morphism. Let $B$ be a submodule of $A$ and $N$ a submodule of $M$ which contains $im \, f$.
Does $F$ preserve the restriction to $B$? More precisely, let $f|_B : B \longrightarrow M$ be the restriction of $f$ to $B$ and $Ff|_{FB} : FB \longrightarrow FM$ the restriction of $Ff$ to $FB$. Is it true that $F(f|_B) = Ff|_{FB}$?
Does $F$ preserve the corestriction to $N$? More precisely, let $f|^N : A \longrightarrow N$ be the corestriction of $f$ to $N$ and $Ff|^{FN} : FB \longrightarrow FN$ the corestriction of $Ff$ to $FN$. Is it true that $F(f|^N) = Ff|^{FN}$?
If we add some conditions to $F$, like $F$ is covariant/contravariant, additive, left/right exact etc... does the answer to 1 or 2 change?
The whole deal with functors is that they respect composition, so the standard maneuver here is to rephrase all of these constructions in terms of composition.
As Randall says in the comments, restriction is just composition with the inclusion $i : B \to A$, so it is still true that $F(f) : F(A) \to F(M)$ can be composed with the map $F(i) : F(B) \to F(A)$ (not necessarily an inclusion unless $F$ preserves monomorphisms, e.g. if $F$ is additive and left exact) to get $F(f \circ i) : F(B) \to F(M)$. This agrees with the restriction of $F(f)$ to $F(B)$ if $F(i)$ is a monomorphism but not in general.
"Corestriction" (I have never heard this name for the thing you're describing) involves lifting $f : A \to M$ to a map $g : A \to N$ such that $j \circ g = f$ where $j : N \to M$ is the inclusion. This gives $F(j \circ g) = F(f) : F(A) \to F(M)$ is the composition of $F(j) : F(N) \to F(M)$ and $F(g) : F(A) \to F(N)$, so again everything is fine if $F(j)$ is a monomorphism but not in general.
As mentioned in 1, $F$ preserves monomorphisms if it is additive and left exact but not in general.