Let $C$ be a category. Let $S\subset C$ be an object. Denote by $C_{/S}$ the slice category over $S$ and by $F$ the forgetful functor from $C_{/S}$ to $C$. Suppose we have morphisms $f:X\rightarrow Z$, $g:Y\rightarrow Z$ in $C_{/S}$. Suppose that $F(f)=F(g)\circ h'$ for a morphism $h':F(X)\rightarrow F(Y)$. Does there necessarily exist a morphism $h:X\rightarrow Y$ such that $F(h)=h'$?
I am primarily interested in the case when $C$ is the category of schemes.
Since the objects in a slice category $C_{/S}$ are morphisms with codomain $S$, let me write $s_X: X \to S$, $s_Y : Y \to S$ and $s_Z : Z \to S$ for the morphisms we need. Now, the morphisms $f : X \to Z$, $g : Y \to Z$ in $C_{/S}$ are really morphisms $s_X \to s_Z$ and $s_Y \to s_Z$, so they satisfy $s_Z \circ f = s_X$ and $s_Z \circ g = s_Y$. Your identity $F(f) = F(g) \circ h$ amounts to $f = g \circ h$ as morphisms in $C$ and the real question is, if we have $s_Y \circ h = s_X$, so that $h$ becomes a morphism $s_X \to s_Y$. In fact, we have $$ s_Y \circ h = (s_Z \circ g) \circ h = s_Z \circ (g \circ h) = s_Z \circ f = s_X $$ and this proves the claim.
You might also want to visualize the diagram with $s_X$, $s_Y$, $s_Z$, $f$, $g$ and $h$ forming a tetrahedron with all but one side commuting and the above shows that the remaining side commutes then, too.