We need the following lemma.
Lemma
Let $K/F$ be a (not necessarily finite dimensional) Galois extension,
$L/F$ an arbitrary extension.
Clearly $KL/L$ is Galois.
Then the restriction map, namely, $\sigma\mapsto \sigma\mid K$ induces
an isomorphism $\psi\colon \mathrm{Gal}(KL/L) \rightarrow \mathrm{Gal}(K/K\cap L)$.
Proof:
We regard $\mathrm{Gal}(KL/L)$ and $\operatorname{Gal}(K/K\cap L)$ as topological groups with Krull topologies.
Clearly $\psi$ is continuous and injective.
Let $H = \psi(\mathrm{Gal}(KL/L))$.
Since $\mathrm{Gal}(KL/L)$ is compact, $H$ is also compact.
Since $\mathrm{Gal}(K/K\cap L)$ is Hausdorff, $H$ is closed.
Clearly the fixed subfield of $K$ by $H$ is $K \cap L$.
Hence $H = \mathrm{Gal}(K/K\cap L)$ by the fundamental theorem of (not necessarily finite dimensional) Galois theory.
This completes the proof.
Now we prove the following proposition with which the OP had a problem.
Proposition
Let $K$ and $L$ be Galois extensions of $F$.
The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces a group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Suppose $K\cap L=F$. Then $\varphi$ is an isomorphism.
Proof.
Since it is clear that $\varphi$ is injective, it suffices to prove that it is surjective.
Let $G_1 = \mathrm{Gal}(K/F), G_2 = \mathrm{Gal}(L/F), G = \mathrm{Gal}(KL/F)$.
By the lemma, given $\sigma_1 \in G_1$, there exists $\sigma \in \mathrm{Gal}(KL/L)$ such that $\sigma\mid K = \sigma_1$. Since $\sigma \in G$ and $\sigma\mid L = 1_L$, $G_1\times 1 \subset \varphi(G)$.
Similarly $1\times G_2 \subset \varphi(G)$.
Hence $G_1\times G_2 = \varphi(G)$.
This completes the proof.
Pick $\beta \in K\setminus F$. Then $1,\beta,\beta^2$ are $F$-linear dependent, but we know that $1,\beta$ are not. Hence $\beta^2+p\beta+q=0$ with $p,q\in F$. Let $\alpha=\beta+\frac12\cdot p$ (this is where the characteristic is used: otherwise we are not allowed to divide by $2$). Now verify that $\alpha^2\in F$.
Best Answer
To prove that the converse is not true, consider the extension $\mathbb{Q}(\sqrt[3]{2},\zeta_{3})$ which is a Galois extension of $\mathbb{Q}$, but $\mathbb{Q}(\sqrt[3]{2})$ is not! (Edit: Take $K=\mathbb{Q}(\sqrt[3]{2})$ and $L=\mathbb{Q}(\zeta_{3})$ and $\mathbb{F}=\mathbb{Q})$ As for the direct statement, we might use splitting fields. Now, if we want to use splitting fields, if $K$ is a splitting field over $F$ for $f(x)$ and $L$ is a splitting field of $g(x)$ over $F$, then $KL$ is a splitting field of...