I am trying to use a well known result of Grothendieck to show that if $S$ is a scheme, and $\mathcal{B}$ is a quasi coherent sheaf of $\mathcal{O}_{S}$-algebras, then there is a relative affine spectrum which is a scheme over $S$.
Denote by $\text{Sch}_{/S}$ the category of schemes over $S$. I am trying to use the result which states that for a functor $$ F: \text{Sch}_{/S} \longrightarrow \text{Set} $$ to be representable, it is necessary and sufficient that it be a Zariski sheaf which is covered by representable open subfunctors.
Grothendieck proves a corollary of this (EGA, 0.4.5.5) which states that if $\{S_{i} \}$ is an affine (or indeed any open) cover of $S$ and $F$ is a Zariski sheaf, then we need only show that the functors $$ F_{i} = F \times_{h_{S}} h_{S_{i}} $$ are representable.
My problem is that I have no idea what this pullback diagram even is. The functor $h_{S}$ is (presumably) the functor which sends any $S$-scheme $(f: X \rightarrow S)$ to the set with one element, namely $\{ f \}$. But then the $h_{S_{i}}$ certainly are not going to be subfunctors of $h_{S}$, and by extension the $F_{i}$ will not be subfunctors of $F$, let alone open subfunctors.
What am I missing here? What is $h_{S}$ actually supposed to be? If I assume that it's an abuse of notation and that $h_{S}$ is really the functor that sends a scheme (not an $S$-scheme) $X$ to the set $\text{Hom}(X, S)$, then what is the morphism of functors $F \rightarrow h_{S}$ supposed to be?
I think what you're looking for is the following. Let $F$ be a sheaf on $X$ and let $\{U_i\}$ be an open cover of $X$. Then, we know by the definition of a sheaf that the diagram
$$0\to F\to \prod_i (\iota_i)_\ast(F\mid_{U_i})\to \prod_{ij}(\iota_{i,j})_\ast(F\mid_{U_i\cap U_j})\qquad (1)$$
is exact where $\iota_i:U_i\hookrightarrow X$ and $\iota_{ij}:U_i\cap U_j\hookrightarrow X$.
Suppose now that $F\mid_{U_i}\cong h_{S_i}$ for some $U_i$-schemes $S_i$. The isomorphisms $(F\mid_{U_i})\mid_{U_j}\cong (F\mid_{U_j})\mid_{U_i}$ give rise to isomorphisms $\varphi_{ij}:(S_i)_{U_i\cap U_j}\xrightarrow{\approx} (S_j)\mid_{U_i\cap U_j}$. We then get a $U$-scheme $S$ (e.g. see Tag01JA). Then, check that $h_S$ sits in the same exact sequence as in $(1)$ (e.g. see the description of the functor of points as in Lemma 25.14.1 of loc. cit.).
I guess technically I only addressed the question on the small Zariski site of $X$, but you can get the claim on the big Zariski site by just noting that if you have any map $f:X'\to X$ you can pull the arguments back to the small Zariski site of $X'$ using the open cover $f^{-1}(U_i)$ and proceeding in the same way.