Exercise. Given sets $E$, $F$, $G$ and functions $g : E \to F$ surjective, and $f : E \to G$. Prove that there's a function $h : F \to G$ such that $f = h \circ g$ if and only if the equality $g\left(x\right) = g\left(y\right)$, with $x, y \in E$, implies the equality $f\left(x\right) = f\left(y\right)$.
My strategy was the following.
Right to left. I assume that $h$ exists. Then if $g(x) = g(y)$, $h(g(x)) = h(g(y))$. Therefore $f(x) = f(y)$ by definition.
Left to right. I assume that the equality $g(x) = g(y)$, with $x,y\in E$, implies the equality $f(x) = f(y)$. Given that $g$ is surjective, $g^{-1}(g(E)) = g^{-1}(F) = E$. Therefore $f(g^{-1}(g(x))) = f(x)$ [Note: I know that $f$ and $g$'s values at any given point $x\in E$ have the same preimages... but I don't know how to state this more formally.]. I can then define $h:\ F\ \longrightarrow\ G$ such that $h(x) = f\circ g^{-1}(x)$, and it verifies the stated condition.
Your proof for right to left is good.
For left to right, note that $g^{-1}(g(E))=E$ for any function $g$, whether or not it is surjective. And because $g$ may not be injective, there is no such thing as $g^{-1}$ and so the expression $$f(g^{-1}\circ g(x)),$$ is not well-defined. Similarly, the definition $$h(x)=(f\circ g^{-1})(x),$$ is faulty because $g^{-1}$ need not exist.
In your note you seem to have the right idea; the fact that $g(x)=g(y)$ implies $f(x)=f(y)$ means that for every $z\in F$ and all $x,y\in g^{-1}(z)$ we have $f(x)=f(y)$, because $$g(x)=z=g(y).$$ Note also that $g^{-1}(z)\subset E$ is non-empty because $g$ is surjective. This means $$h(z):=f(x),\qquad\text{ where } x\in g^{-1}(z),$$ is a well-defined function $h:\ F\ \longrightarrow\ G$, and it clearly satisfies $f=h\circ g$.