You are confused about the Galois correspondence - normal subgroups $H$ of the Galois group $\text{Gal}(E/F)$ correspond to normal extensions $E^H/F$, where $E^H$ denotes the subfield of $E$ fixed by $H$. Note that $E$ being the splitting field of a polynomial in $F$ does not guarantee that $E/F$ is Galois. This is due to the fact that $E/F$ is Galois only when it is normal (i.e., is a compositum of some splitting fields) and separable.
However, I imagine the statement you intended was:
If $E/F$ is normal, then $$H\triangleleft \text{Aut}(E/F) \iff E^H/F \text{ is a normal field extension.}$$
This is actually still true. It can be regarded as a salvaging of the Fundamental Theorem of Galois Theory in the case that $E/F$ is not necessarily separable. Here is my reasoning: Let $E/F$ be normal and let $G=\text{Aut}(E/F)$. Then $E^G/F$ is purely inseparable, and $E/E^G$ is separable. We have that $\text{Aut}(E/E^G)=\text{Aut}(E/F)$. Because $E/F$ is normal, we have that $E/E^G$ is normal and hence $E/E^G$ is Galois, and therefore a normal subgroup $H\triangleleft\text{Aut}(E/E^G)=\text{Aut}(E/F)$ corresponds to a normal extension $E^H/E^G$. It is known that if $C\subseteq B\subseteq A$ is a tower of field extensions and $A/C$ is normal and $B/C$ is purely inseparable, then $A/B$ is normal. Thus $E^H/F$ is normal.
Conversely, given a normal subextension $L/F$ of $E/F$ that is the fixed field $L=E^H$ of some subgroup $H\subseteq\text{Aut}(E/F)=\text{Aut}(E/E^G)$, then $L$ in fact contains $E^G$, and $L/F$ normal implies $L/E^G$ normal, hence the subgroup of $G$ that fixes $L$, namely $H$, is normal in $G$.
Please take note of this question: it is not an obvious one!
However, constructing a counterexample is evading me. Here is a try:
Let $\mathbb{F}_p$ be a finite field where $3\nmid p-1$ (so that there are no cube roots of unity), let $F=\mathbb{F}_p(T)$, let $f=x^{3p}-T\in F[x]$, let $E$ be the splitting field of $f$ over $F$. Then $E=F(\sqrt[3p]{T},\sqrt[3]{1})$. This has as a subfield $M=F(\sqrt[3]{T})$, which is separable, but not normal, over $F$. EDIT: Nevermind, this doesn't work. $M$ isn't the fixed field of a normal subgroup of $\text{Aut}(E/F)$.
This is a standard fact, part of the Fundamental Theorem of Galois Theory.
Assuming separability throughout, let’s take this as a definition of $E$ being Galois over a field $k$: that every $k$-morphism $\varphi$ of $E$ into a separably closed extension $\Omega$ of $E$ actually sends $E$ into $E$.
So let’s suppose that $F_2$ is Galois over $F_0$ and that $\text{Gal}(F_2/F_1)
$ is a normal subgroup of $\text{Gal}(F_2/F_0)$. Now let $\varphi$ be an $F_0$-morphism of $F_1$ into $\Omega$. It can be extended to $\bar\varphi\colon F_2\to\Omega$, by a standard theorem, and by the hypothesis that $F_2$ is Galois over $F_0$, it sends $F_2$ to $F_2$, and thus is an element of $\text{Gal}(F_2/F_0)$. I want to show that $\bar\varphi(F_1)\subset F_1$, by showing that any element $\bar\varphi(x))$ of this field is fixed under $\text{Gal}(F_2/F_1)$. Now let $g$ be any element of this group. I want $g(\bar\varphi(x))=\bar\varphi(x)$. But by normality of the subgroup, $g\circ\bar\varphi=\bar\varphi\circ g'$, for an (perhaps other) element $g'$ of the subgroup. But since $g'$ leaves $x$ fixed, we have $g(\bar\varphi(x))=\bar\varphi(x))$, as desired.
This argument is probably unnecessarily wordy; I’m sure you can improve it.
Best Answer
If $L/K$ is a Galois extension, and $M$ is any intermediate field, then $L/M$ is also Galois, and ${\rm Gal}(L/M)$ is a subgroup of ${\rm Gal}(L/K)$, by the fundamental theorem of Galois theory.
If it so happens that $M/K$ is a Galois extension, then $Gal(L/M)$ is normal in ${\rm Gal}(L/K)$ (and vice versa), and we also have $${\rm Gal}(L/K)\,/\,{\rm Gal}(L/M)\cong{\rm Gal}(M/K).$$