Let $X$ be a scheme and $\mathscr{R}$ a quasi-coherent $\mathscr{O}_{X}$-algebra. Let
$$ F: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_X-Alg)}(\mathscr{R},f_{*}\mathscr{O}_T) $$
be a functor. This question is adressed to representability of this functor $F$ and the only answer contains some remarks I not understand.
We introduce there an affine cover $\bigcup_i U_i$ of $X$ and consider the functors
$$ F_i: (Sch/X)^{opp} \rightarrow (Sets), \hspace{1cm} (f:T \rightarrow X) \mapsto \text{Hom}_{(\mathscr{O}_{X}-Alg)}((u_i)_*(\mathscr{R}\vert_{U_i}),(f_{*}\mathscr{O}_T)) $$
where $u_i: U_i \hookrightarrow X$ is the canonical inclusion.
We observe that we have for every $i$ a canonical morphism $ \alpha_i:\mathscr{R}\to (u_i)_*(\mathscr{R}|_{U_i})$ induced by $u_i$. This induces a morphism of functors $\alpha_i^*:F_i\to F$. It is claimed that $\alpha_i^*$ is a monomorphism.
Moreover it is also claimed there that this description implies immediately that for any $X$-scheme $f: X \to X$ we must have that $F_i(T) \neq \emptyset$ if and only if $f: T \to X$ factors through $U_i$.
Questions: Why is $\alpha_i^*:F_i\to F$ a monomorphism of functors? Or equivalently why for every $X$-morphism $f: T \to X$ there is an inclusion $F_i(T) \subset F(T)$?
Why from the construction of $F_i$ is follows that $F_i(T) \neq \emptyset$ if and only if $f: X \to X$ factors through $U_i$?