I am trying to understand this theorem from Hungerford's abstract algebra textbook.
So the part that I am not completely sure of is why $K/E$ is a separable field extension. I think the reason is because it would contradict the fact that $p(x)$ divides $f(x)$ because $p(x)$ is the minimal polynomial of $u$ in $E[x]$ and since $f(x)$ is also in $E[x]$ and contains $u$ as a root. This would imply that $p(x)$ also divides $f(x)$ in the splitting field of $p(x)$ over $E$. But if $p(x)$ have roots of multiplicity higher than 1, then this would not be possible.
Another question I have is why is $f(x)$ not in fact equal to the minimal polynomial $p(x)$. The reason is because $f(x)$ is simply linear factors of the possible roots of $p(x)$ that are mapped to by $H$ on $u$. However, this does not mean that those are in fact all the roots of $p(x)$. The typical example given is usually $Q(2^{1/3})/Q$. The only automorphisms of $Q(2^{1/3})/Q$ that fixes $Q$ is the identity, because the other roots of $x^3-2$ are complex numbers. So by that logic, we should have degree of $p(x)\geq f(x)$. On the other hand, since $p(x)$ divides $f(x)$, we must have the degree of $p(x)\leq f(x)$, and since they are both monic polynomials, they should be equal right?
To answer the first question, your understanding is entirely correct. To rephrase: if an element in a field extension satisfies any separable polynomial, its minimal polynomial is a factor of a separable polynomial, hence necessarily separable as well.
To answer the second question, yes, $f$ is in fact the minimal polynomial of $u$. This is not necessary to know for the argument to work, but we may as well appreciate this fact. By construction, $f$ is monic. Let $g\in E[x]$ be any polynomial, such that $g(u)=0$. For any $\sigma\in H$, we have that $g(\sigma(u))=\sigma(g(u))=\sigma(0)=0$, since $\sigma$ is a field automorphism. Since the $H$-orbit of $u$ is precisely $\{u_1,\dotsc,u_t\}$, it follows that $u_1,\dotsc,u_t$ are all zeros of $g$, i.e. $x-u_1,\dotsc,x-u_t$ all divide $g$. Since they are coprime, it follows that $f(x)=(x-u_1)\cdot\dotsc\cdot(x-u_t)$ divides $g$. Thus, $f$ satisfies the defining properties of the minimal polynomial of $u$.
Your example is misguided, because $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not a Galois extension. Indeed, the fact that the minimal polynomial of $2^{1/3}$ does not split in $\mathbb{Q}(2^{1/3})$ is precisely observing that the extension is not normal.