This is false in general. Let me begin by offering an example from the side of finite groups (the Galois correspondence is inclusion reversing, so...). Let $G=D_8$ be the group of symmetries of a regular octagon. We know that $G$ is generated by a rotation $r$ of order $8$, and a reflection $s$ of order $2$ such that $srs=r^{-1}$, and it follows that $sr^i=r^{8-i}s$ for all integers $i$. It follows that $|G|=16$ and that $r^4$ is the only non-trivial element in the center of $G$. Geometrically this is obvious, because $r^4$ is the rotation by the angle $\pi$, i.e. scalar multiplication by $-1$.
Inside $G$ we have two subgroups $H_1=\{1,s,r^4, sr^4\}$ and $H_2=\{1,sr,r^4,sr^5\}$. Both of these are generated by reflections w.r.t. to a pair of orthogonal lines - one pair gotten by rotating the other by the angle $\pi/8$. Geometrically: one pair of the the orthogonal lines connects two pairs of antipodal vertices of the octagon, the other pair connects antipodal midpoints of edges. Anyway, the key observations are the following:
- $\langle H_1,H_2\rangle=G$. This is because $r=s(sr)$ is in the subgroup generated by both $H_1$ and $H_2$. In other words no proper subgroup of $G$ contains both $H_1$ and $H_2$.
- $H_1\cap H_2=\{1,r^4\}$ is non-trivial.
Translating this to field extensions forming a counterexample is straightforward.
Let $E/F$ be a Galois extension with Galois group $G$. Let $K=\operatorname{Inv}(H_1)$
and $L=\operatorname{Inv}(H_2)$. The intersection $L\cap K$ is then invariant under $H_1$ and $H_2$, hence under all of $G$, so $L\cap K=F$. OTOH the compositum $LK$ is the smallest field containing both $L$ and $K$, and thus the field of invariants of $H_1\cap H_2$. Here $[K:F]=[G:H_1]=4$, $[L:F]=[G:H_2]=4$, $[E:F]=|G|=16=4\cdot4$, but
$[LK:F]=[G:(H_1\cap H_2)]=8<16$.
Translating everything to the group side by an application of the Galois correspondence was the key. After that it was just tinkering, trial and error.
Observe that it suffices to find a finite group $G$ together with two subgroups $H_1$ and $H_2$ such that i) $G=\langle H_1, H_2\rangle$, ii) $|G|=[G:H_1]\cdot[G:H_2]$ and ii) $|H_1\cap H_2|>1$.
I’m going to interpret your question as asking whether $\mathrm{Aut}(K_1K_2/F) = \mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)$ if $K_1$ and $K_2$ are extensions of $F$ such that $K_1\cap K_2=F$, where for fields $L\subseteq M$,
$$\mathrm{Aut}(M/L) = \{\sigma\in\mathrm{Aut}(M)\mid \sigma|_L=\mathrm{id}_L\},$$
and hence equals the Galois group in the special case where $M$ is Galois over $L$.
Consider $F=\mathbb{Q}$, $K_1=\mathbb{Q}(\sqrt[3]{2})$, and $K_2=\mathbb{Q}(\omega)$, where $\omega$ is a complex cubic root of unity; that is, a root of $x^2+x+1$.
Then $K_2$ is Galois over $\mathbb{Q}$ and $\mathrm{Gal}(K_2/F)$ is cyclic of order $2$. On the other hand, $\mathrm{Aut}(K_1/F)$ is trivial, because $\sqrt[3]{2}$ is the only real cubic root of $2$, $K_1$ is contained in $\mathbb{R}$, and an automorphism of $K_1$ must send $\sqrt[3]{2}$ to a cubic root of $2$. Thus,
$$\mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)\cong\{e\}\times C_2 \cong C_2$$
is cyclic of order $2$.
However, $K_1K_2$ is the splitting field of $x^3-2$, which has degree $6$ over $F$. In particular, $\mathrm{Gal}(K_1K_2/F)$ has order $6$ (and is isomorphic to $S_3$) so we get neither an isomorphism with, nor a realization as a subgroup of the direct product.
Best Answer
No! Let $k=\mathbb{Q}$, and consider $K_1=\mathbb{Q}(\sqrt{2}), K_2=\mathbb{Q}(\sqrt{3})$ and $K_3=\mathbb{Q}(\sqrt{6})$. We have that $K_1 \cap K_3 = K_2 \cap K_3=\mathbb{Q}$, and so their compositum is also $\mathbb{Q}$. On the other hand, $K_1K_2 = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ contains $\sqrt{6}$, and so $$ K_1K_2 \cap K_3 = \mathbb{Q}(\sqrt{6}) \supsetneq \mathbb{Q} = (K_1 \cap K_3)(K_2 \cap K_3). $$