If $E\supseteq K\supseteq F$ and $E$ is a normal extension of $F$, then $E$ is a splitting field over $F$, so $E$ is a splitting field over $K$, and therefore $E$ is a normal extension of $K$. It need not be the case that $K$ is a normal extension of $F$. The next lemma characterizes those $K$ that are normal extensions of $F$.
Lemma 1.4.1 If $E\supseteq K\supseteq F$, and $E$ is a normal extension of $F$, then $K$ is a normal extension of $F$ if and only if for all $\sigma\in G(E/F)$, $\sigma(K)\subseteq K$.
Proof. Suppose that for all $\sigma\in G(E/F)$, $\sigma(K)\subseteq K$. Let $K=F(a)$, then for all $\sigma\in G(E/F)$, $\sigma(a)\in K$. Let $p(x)=\prod_{\sigma\in G(E/F)}(x-\sigma(a))$. As in the proof of theorem 1.3.4, $p(x)\in F[x]$ and $K$ is the splitting field of $p(x)$ over $F$. Hence $K$ is a normal extension of $F$.
For the converse, suppose $K$ is a normal extension of $F$ and $K=F(a)$. Let $p(x)=\prod_{\sigma\in G(K/F)}(x-\sigma(a))$—note the $K$! Then just as in the proof of theorem 1.3.4, $K$ is the splitting field of $p(x)$ over $F$, and $p(a)=0$. Then for every $\sigma\in G(E/F)$ (note the $E$!), $0=\sigma(p(a))=p(\sigma(a))$, so $\sigma(a)$ is a root of $p$. Since all roots of $p$ are in $K$, $\sigma(a)\in K$.
The element $a$ has a minimal polynomial over $F$, with some degree $n$. By theorem 19.3 in Gallian, every element $b$ of $K=F(a)$ can be written as $b=c_0+c_1 a+c_2 a^2+\cdots+c_{n-1}a^{n-1}$ for some $c_i\in F$. Thus $$\sigma(b)=\sum_{i=0}^{n-1}c_i (\sigma(a))^i\in K,$$ so $\sigma(K)\subseteq K$. $\qed$
Theorem 1.4.2 (Fundamental Theorem of Galois Theory) Suppose $E$ is a splitting field over $F$ and $E\supseteq K\supseteq F$. Recall the two functions: $$\eqalign{ f(H)&= E_H, \hbox{ if $H\le G(E/F)$}\cr g(K)&= G(E/K)\cr}$$
1. $f$ is a bijection, with inverse $g$, from the subgroups of $G(E/F)$ onto the subfields of $E$ that contain $F$. In other words, $K=E_{G(E/K)}$ and $H=G(E/E_H)$.
2. $[E\,\colon K]=|G(E/K)|$ and $[K\,\colon F]=|G(E/F)|/|G(E/K)|$.
3. $K$ is a normal extension of $F$ iff $G(E/K)$ is a normal subgroup of $G(E/F)$.
4. If $K$ is a normal extension of $F$, then $G(K/F)\cong \gmod {G(E/F)}/G(E/K).$.
Proof. 1) Suppose $E\supseteq K\supseteq F$. Since $E$ is a splitting field over $F$, it is also a splitting field over $K$. Hence $E$ is a normal extension of $K$ and so by definition of normal, $K=E_{G(E/K)}$. By theorem 1.3.2, if $H\le G(E/F)$ then $H=G(E/E_H)$.
2) Suppose $E\supseteq K\supseteq F$. Again by theorem 1.3.2, $[E\:F]=|G(E/F)|$ and $[E\:K]=|G(E/K)|$. Then $|G(E/F)|=[E\:F]=[E\:K][K\:F]=|G(E/K)|[K\:F]$, so $[K\,\colon F]=|G(E/F)|/|G(E/K)|$.
3) By lemma 1.4.1, $K$ is a normal extension of $F$ iff $$\eqalignno{ &\forall\sigma\in G(E/F)\forall t\in K \bigl(\sigma(t)\in K\bigr).& (1.4.1)\cr }$$ Because $E$ is a normal extension of $K$, the fixed field of $G(E/K)$ is $K$, so condition 1.4.1 is equivalent to $$\eqalignno{ &\forall\sigma\in G(E/F)\forall\tau \in G(E/K) \forall t\in K \left(\tau\bigl(\sigma(t)\bigr)=\sigma(t)\right).& (1.4.2)\cr }$$ Condition 1.4.2 is equivalent to $$\forall\sigma\in G(E/F)\forall\tau \in G(E/K) \forall t\in K (\sigma^{-1}(\tau(\sigma(t)))=t), $$ which means $\sigma^{-1}\tau\sigma$ fixes $K$, or $\sigma^{-1}\tau\sigma\in G(E/K)$. By theorem 9.1 in Gallian, this is equivalent to $G(E/K)\mathrel{\triangleleft} G(E/F)$.
4) Suppose $K$ is a normal extension of $F$. Let $\sigma \in G(E/F)$, and denote the restriction of $\sigma$ to $K$ by $\overline\sigma$. By lemma 1.4.1, $\sigma(K)\subseteq K$, so $\overline\sigma$ is an isomorphism to a subfield of $K$. Viewing $K$ as a finite dimensional vector space, we have an isomorphism to a subspace of $K$ with the same dimension as $K$, so $\overline\sigma(K)=K$ and $\overline\sigma\in G(K/F)$. Since $\overline{\sigma\circ\tau}=\overline\sigma\circ\overline\tau$, the mapping $\sigma\mapsto\overline\sigma$ is a homomorphism, $\phi$, from $G(E/F)$ to $G(K/F)$.
The kernel of $\phi$ consists of those $\sigma$ such that $\overline\sigma$ is the identity automorphism on $K$, that is, such that $\sigma$ fixes $K$. Thus, the kernel is precisely $G(E/K)$. By part (3), $G(E/K)\mathrel{\triangleleft} G(E/F)$, and by the First Isomorphism Theorem for Groups, $\gmod{G(E/F)}/G(E/K).$ is isomorphic to the range of $\phi$.
We know that $$\left|G(K/F)\right|=[K\:F]=\left|\gmod{G(E/F)}/G(E/K).\right|= \left|\phi\bigl(G(E/F)\bigr)\right|. $$ Thus the range of $\phi$ is a subgroup of $G(K/F)$ with the same size as $G(K/F)$, so the range is $G(K/F)$, which finishes the proof. $\qed$