If $\Phi$ is injective, $X_1$ normal and Hausdorff and $X_2$ compact, then $F$ is surjective

Good ideas here, but re-order them and go for the simplest approach first:

Suppose $F[X_2] \neq X_1$; we will derive a contradiction.

So pick $x_1 \in X_1\setminus F[X_2]$, to witness the non-equality. Note that $F[X_2]$ is compact ($X_2$ is compact and $F$ is continuous) and so $F[X_2]$ is closed in $X_1$ (as $X_1$ is Hausdorff).

Then apply your idea and find a Urysohn function $g: X_1 \to [0,1]$ such that $g(x) = 0$ for all $x \in F[X_2]$ and $g(x_1)= 1$ (using that $\{x_1\}$ is also closed, from Hausdorffness, and normality of $X_1$ of course).

Then $\Phi(g)$ is the function that is identically $0$ on $X_2$, call it $\textbf{0}_{X_2}$ say, as

$$\forall x \in X_2: \Phi(g)(x)= g(F(x)) = 0 \text{ because } F(x) \in F[X_2] \text{ and the construction of } g$$

but as $\Phi(\textbf{0}_{X_1}) = \textbf{0}_{X_2}$ as well and $g \neq \textbf{0}_{X_1}$ we have a contradiction with the injectivity of $\Phi$.