Above I have posted the question (and Exercise 4.1.9 which the question references in its hint). I am a bit confused by the hint. It references the Galois group of L over F, but L is just a splitting field and not necessarily Galois
Follow the hint. Suppose that $f(x) = p_1(x)... p_m(x)$ over $K$ with each $p_i$ irreducible over $K$. Since $L$ is the splitting field for $f$, $f$ factors completely into linear terms. Since a polynomial ring over any field is a UFD, this implies each $p_i(x)$ factors over $L$. Namely $p_i(x) \in L[x]$. Hence $p_i(x) \in (L \cap K)[x]$. Since clearly $p_i(x)$ cannot be irreducible over $K$ and reducible over $L \cap K$ (as the latter is contained in the former), this implies $f$ has the same factorization over $K$ as $L \cap K$.
If L/F were Galois, then so is $L/L\cap K$ by the Fundamental Theorem of Galois theory and $H:= Gal(L / L \cap K) \leq Gal(L/F) := G$. Since each $p_i$ is irreducible in $L \cap K$ but has its roots in $L$, we know that any element $\sigma \in H$ just permutes the roots of $p_i$. Hence, the roots of the individual factors $p_i$ are the orbits under the action of $H$ on the set of roots of $f$.
If I could prove H was a normal subgroup of $G$ (Fundamental Theorem of Galois theory seemed promising here?), I could then directly apply Exercise 4.1.9a to say all the orbits have the same size (e.g. all the $p_i$ have the same number of roots and hence the same degree).
If WLOG $\alpha \in L$ is a root of $p_1$, then as $p_1$ is irreducible it is the minimal polynomial of $\alpha$. Hence $deg(p_i) = deg(p_1) = [K(\alpha):K]$. So just let $d = [K(\alpha):K]$ and $m = n/d$.
Really, my main problems are with $L/F$ being Galois and $H$ being normal. Thanks in advanced!