Let $X\overset{f'}\to S$ and $T\overset{g}\to S$ be schemes over $S=Spec(k)$ for a field $k$. Further, let $X_T = X \underset{S}\times T$ denote the fiber product over $S$, and $f$ the induced map $X_T\to T$. I run into the claim $$ \Gamma(X_T,\; \mathcal{O}_{X_T}) \cong \Gamma(T,\; f_*\mathcal{O}_{X_T}), $$ and I shoved to myself that this is the case by the identitfications of functors $$ \Gamma(X_T, \square) \cong (g\circ f)_* \square \cong g_* \;f_* \square \cong \Gamma(T, f_*\square). $$
I have two questions:
Is there some kind of adjunction formula or well-known result that implies the above claim?
Do I really need the hypothesis $S=Spec(k)$, or does this work over an arbitrary scheme $S$?
For the first question, this is clear by definition of $f_\ast$:
The sheaf $f_\ast \mathcal O_{X_T}$ is given by associating to an open subset $U$ of $T$ the set $\mathcal O_{X_T}(f^{-1}(U))$. So a global section of $f_\ast \mathcal O_{X_T}$ is just given by $$\Gamma(T,f_\ast \mathcal O_{X_T})=\mathcal O_{X_T}(f^{-1}(T))=\mathcal O_{X_T}(X_T)=\Gamma(X_T,\mathcal O_{X_T})$$
For the second question, this clearly works over any base scheme. You don't even have to talk about fiber products.