Where $\Gamma$ is the functor that takes a topological space and a sheaf of module and gives the section of that sheaf at that space. The right derived functor is taken as if $X$ is fixed.
This is from Illusie chapter one, the part about sheaf cohomology.
Thanks.
This is a general statement about right derived functors of left exact functors : if $F: C\to D$ is such a functor between abelian categories, and $C$ has enough injectives, then for any $X \in C$, we have $H^0(RF(X)) \cong F(X)$.
The proof is easy : take an injective resolution $X\to I^\bullet$, then $RF(X) \cong F(I^\bullet)$ and so $H^0(RF(X)) \cong H^0(F(I^\bullet))$. Now $0\to X\to I^0\to I^1$ is exact, so that $0\to F(X) \to F(I^0)\to F(I^1)$ is as well (by left exactness of $F$), and so $F(X) \cong \ker(F(I^0)\to F(I^1)) = H^0(F(I^\bullet))$.
This isomorphism is natural, and the precise proof of that fact will depend on how precisely you set up the definition of $RF$, but in any case it won't be hard to prove.