The Galois group of a composite of Galois extensions

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.

I think this solution will give more general intuition.

Let $\varphi : Gal(KL/F) \to Gal(K/F) \times Gal(L/F)$ which sends $\sigma \to (\left.\sigma\right|_{K}, \left.\sigma\right|_{L})$.

Lemma: $Gal(KL/F)$ is isomorphic to the subgroup $H = \{(\sigma, \tau) \}$ of $Gal(K/F) \times Gal(L/F)$ where $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$. So $\varphi$ is surjective if and only if $H = Gal(K/F) \times Gal(L/F)$, and this happens if and only if $K \cap L = F$.

Proof. $\varphi$ is clearly injective (any element in kernel must fix $K$ and $L$, therefore, it fixes $KF$ which implies the identity). Let $A = img(\varphi)$. It's obvious that $A \subset H$. To prove the reverse, not that $KL$ is Galois over $F$ and $K \cap L$ is a subfield containing $F$. Then any embedding from $K \cap L$ to $\bar{F}$ is of the form $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ (generally, given field $F$ with Galois extension $K$, if $E$ is subfield of $K$ containing $F$, then any embedding of $E$ which fixes $F$ is of the form $\left.\phi\right|_{E}$ for some $\phi \in Gal(K/F)$. This is a nice property, and I'll let you play with it!). If $(\phi, \tau) \in H$, then $\left.\sigma\right|_{K \cap L} = \left.\tau\right|_{K \cap L}$ and $\left.\sigma\right|_{K \cap L} =\left.\tau\right|_{K \cap L}$ is actually of them form $\left.\phi\right|_{K \cap L}$ for some $\phi \in Gal(KL/F)$ which implies $H \subset A$, hence $H=A$.

Hope this helps!