Let $f:X\longrightarrow Y$ be a continuous mapping of topological spaces. Let $F$ be an abelian sheaf on $X$. We define $R^i f_{\star} F$ as the $i$-homology sheaf of the complex $f_{\star}(D^{\star}(F))$, where $(D^{\star}(F))$ is the usual canonical sheaf resolution used to define cohomology.
I'm really struggling trying to understand the following statement
One way to compute $R^i f_{\star} F$ is to note that it is the sheaf associated to the presheaf $V\rightarrow H^i(f^{-1}V,F)$ for any open subset $V$ of $Y$.
Even though this sounds incredibly intuitive to me, I have no idea how to prove it.
This is not yet a full answer, but try it as an exercise: For any open $V \subset Y$ you have $$f_*(D^\bullet F)(V) = D^\bullet F(f^{-1}V).$$ So if you take cohomology at that step, you obtain $$H^k(f_*(D^\bullet F)(V)) = H^k(f^{-1} V, F).$$ Now it remains to show that whenever $C^\bullet$ is a cocomplex of sheaves, the cohomology sheaves $H^k(C^\bullet)$ are the sheaves associated to the presheaf $$V \mapsto H^k(C^\bullet (V)).$$