Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x]

It becomes easy if you know about integrality over rings. (Here and in the following, "ring" always means "commutative ring with $1$".)

Since $K$ is a finite extension of $F$, we see that $k\in K$ is integral over $F$ for every $k\in K$. Thus, $k\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$. Hence, $kx^i\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$ and $i\in\mathbb N$ (since $k$ and $x^i$ are both integral over $F\left[x\right]$, and the product of two integral elements is integral). Hence, $f\left(x\right) \in K\left[x\right]$ is integral over $F\left[x\right]$ (since $f\left(x\right)$ is a sum of elements of the form $kx^i$ for $k\in K$ and $i\in\mathbb N$, and since the sum of integral elements is integral). Now, all we need to prove is the following fact:

(1) If $ A\subseteq B$ is a ring extension, and $ u\in B$ is integral over $ A$, then there exists a nonzero $ v\in B$ such that $ uv\in A$.

Proof of (1). Let $ n$ be the smallest positive integer such that there exists a monic polynomial $ P\in A\left[Y\right]$ of degree $ n$ satisfying $ P\left(u\right) = 0$. (Such an $n$ exists since $u$ is integral over $A$.) Write the polynomial $P$ in the form $ P\left(Y\right) = \sum\limits_{i = 0}^{n - 1}a_iY^i + Y^n$ with all $a_i$ lying in $A$. Then, set $ v = \sum\limits_{i = 1}^{n - 1}a_iu^{i - 1} + u^{n - 1}$. Then, $uv = \sum\limits_{i = 1}^{n - 1}a_iu^{i} + u^{n} = \underbrace{\sum\limits_{i = 0}^{n - 1}a_iu^{i} + u^{n}}_{ = P\left(u\right) =0} - a_0 = - a_0 \in A$. Also, $ v\neq 0$ follows from the minimality of $ n$. Thus, (1) is proven.

[I have copied some of this from one of my AoPS posts.]

Assume that the extension is Galois. By assumption, there are only finitely many automorphisms of $K$ that fix $F$. The polynomial $$ \omega(x) = \prod_{\sigma \in \mathrm{Aut}(K/F)} \sigma(f(x)) $$ is a polynomial such that $f(x) \, | \, \omega(x)$ and $\omega$ is fixed by every automorphism of $K/F$, because $\mathrm{Aut}(K/F)$ is a group, so that when one tries to apply an automorphism on $\omega(x)$, the conjugate factors get permuted... hence all the coefficients of $\omega$ lie in the fixed field $F$. You can choose $g(x) = \omega(x) / f(x)$.

Hope that helps,