Let $\mathscr{C}$ be the category of schemes (that is the category of schemes over $\mathrm{Spec}(\mathbb{Z})$), $X,Y$ be two schemes. Then we can have two functors: $\mathscr{C}^{op}\to \{\mathrm{Sets}\}$, one is $h_{X\coprod Y}$ ($X\coprod Y$ is the disjoint union (or coproduct) of $X,Y$), where $h_{X\coprod Y}(T)$ is the $Mor(T, X\coprod Y)$ for all scheme $T$. The other is $h_{X}\coprod h_{Y}$ defined by $(h_{X}\coprod h_{Y})(T)=Mor(T,X)\coprod Mor(T,Y)$. I think these two functors are not isomorphic.
For example, if $T=X\coprod Y$, then we have $\mathrm{id}: T\to X\coprod Y$ which belongs to $h_{X\coprod Y}(T)$,but we could not find an element in $(h_{X}\coprod h_{Y})(T)$ which corresponds it. Hence there is no canonical isomorphism between $h_{X\coprod Y}(T)$ and $(h_{X}\coprod h_{Y})(T)$. Thus these two functors are not isomorphic. Am I right?
You are correct, these two functors are generally not equivalent. Your argument is on the right track, but needs a bit of polishing to make precise. Below is one way to make this precise.
Let $X$ and $Y$ be nonempty schemes. Then the identity morphism $$1_{X\sqcup Y} \in \mathrm{Mor}(X \sqcup Y, X \sqcup Y)$$ is surjective. On the other hand, there can be no surjective morphism in $$\mathrm{Mor}(X \sqcup Y, X) \sqcup \mathrm{Mor}(X \sqcup Y, Y)$$ since the image of any such morphism must land completely in $X$ or completely in $Y$.