Let $X$ be a Noetherian scheme and $Sh(X)$ denote the category of sheaf of abelian groups on $X$.Let $x\in X$ be a closed point. Let $\mathfrak F : Sh(X) \to Sh(X)$ be an additive left exact functor such that the assignment $\mathfrak G: Sh(X) \to Ab$ sending $\mathcal F\to \mathfrak F(\mathcal F)_x$ is also an additive left exact functor.
Then , is it true that $(R^j \mathfrak F)(\mathcal F)_x \cong (R^j\mathfrak G)(\mathcal F)$ , as abelian groups , for every $\mathcal F \in Sh(X)$ ?
If this is not true in general, then is it at least true if we also assume $\mathfrak F(\mathcal F)_x=\mathfrak F(\mathcal F)(X), \forall \mathcal F\in Sh(X)$ ...?
Here $R^j$ denotes the $j$-th right derived functor.
Let $A,B,C$ be abelian categories with enough injectives (so that left derived functor exists etc). Let $F:A \to B$ be a left exact additive functor between two abelian categories and let $G:B \to C$ be an exact additive functor. To translate to your notation $F$ is your $\mathfrak{F}$, I write $G$ for the $(-)_x$ functor (which is exact) and then $G \circ F$ is what you denote by $\mathfrak G$.
In this case $RG=G$ where $RG$ is the total derived functor, and the required result will follow is we can show that $R (G \circ F) = RG \circ RF$. This is not always true, but holds when $F(I)$ is acyclic for all injective objects $I$ of $A$ (this is https://stacks.math.columbia.edu/tag/015L), i.e., that $R^i G F(I)=0$ for $i>0$. In our situation $G$ is exact and so this condition is automatically satisfied, meaning that we get
$$ R(G \circ F)=RG \circ RF = G \circ RF, $$ and so $$ R^j (G \circ F) = G \circ R^j F. $$