Assume that $G = \operatorname{Gal}(L/\Bbb{Q}), H = \operatorname{Gal}(K/\Bbb{Q})$.
Can we construct a field extension of $\Bbb Q$ from $L$ and $K$ such that $G \times H$ is its Galois group?
For $L \cap K = \Bbb Q$, the field extension we need is just $LK$. What if $L \cap K = M, \Bbb Q < M$?
Are there corresponding results for fields other than $\Bbb Q$?
On
Perhaps a $p$-adic approach could shed some light on some cases where the OP problem has a negative answer. Fix an odd (for simplicity) prime $p$. Given a number field $F$, let $S$ be a finite set of primes of $F$ containing the set $S_p$ of primes above $p$, and consider the maximal pro-$p$-extension $F_S$ of $F$ which is unramified outside $S$, with $G_S(F):=Gal(F_S/F)$. It follows from global CFT that the group $G_S(F)^{ab}$ (obvious notation) is a $\mathbf Z_p$-module of finite type, and Leopoldt's conjecture (proved for abelian fields) gives its $\mathbf Z_p$-structure: (1) $G_S(F)^{ab}\cong \mathbf Z_p^{1+c} \times T_S(F)$, where $T_S(F)$ denotes the (finite) $\mathbf Z_p$-torsion submodule and $c$ is the number of complex primes of $F$. Let us keep the OP notations $K, L, G=Gal(K/\mathbf Q), H=Gal(L/\mathbf Q)$.
1) If $F=\mathbf Q$, then $T_{S_p}(\mathbf Q)=0$ and the decomposition (1) reads $G_S(\mathbf Q)^{ab}\cong \mathbf Z_p \times T_S(\mathbf Q)$, where $\mathbf Z_p$ is the Galois group of the cyclotomic $\mathbf Z_p$-extension of $\mathbf Q$ (which is $\mathbf Q_{S_p}$), and $T_S(\mathbf Q)$ is the product over the primes in $T=S-S_p$ of the local $p$-primary groups of roots of unity. This gives explicitly the structure of the finite quotients of $G_S(\mathbf Q)^{ab}$ and provides a negative answer to the OP question when it is restricted to abelian $p$-groups. Indeed, we can always enlarge $S$ in order to include the primes of $\mathbf Q$ which ramify in $K$ and $L$, and then fix an extension $E/\mathbf Q$ with Galois group $\cong T_S(\mathbf Q)$. Then $K$ (and $L$ as well) is the compositum of a sub-extension of $E$ and a sub-extension of $\mathbf Q_{S_p}$; and so will be $K.L$ . But $G\times H$ cannot be a quotient of $G_S(\mathbf Q)^{ab}$ as described above, even if $K\cap L=\mathbf Q$. This is analogous to the example over a finite field given by @Quiaochu Yuan.
2) The non abelian case is perhaps more "visible" over a general number field $F$. Let $S, S_p, T$ be as before (relatively to $F$). Modulo Leopoldt's conjecture, the following can be proved. Let us say that $F$ is $p$-rational if $T_{S_p}(F)=0$, which is equivalent to saying that $G_{S_p}(F)$ is a free pro-$p$-group. Example: $\mathbf Q$ is $p$-rational. Let $\tilde F$ be the compositum of all the $\mathbf Z_p$-extensions of $F$, so that $\tilde F$ is the fixed field of $T_S(F)$; the set $S$ is called primitive if the Frobenius automorphisms $\phi_w$ of $Gal(\tilde F/F)$ for all $w\in T$ generate a free $\mathbf Z_p$-submodule $Gal(\tilde F/F)$ of rank equal to $\mid T \mid$. If $F$ is $p$-rational and $S$ is primitive, then (2) $G_S(F) \cong G_{S_p}(F) * (*_{w\in T} G_w)$, where * denotes the free product in the category of pro-$p$-groups and $G_w$ is the inertia subgroup of $G_S(F)$ relative to $w$ ((2) can be viewed as the non abelian lift of (1)). This theorem shows that: (i) Any $p$-group whose minimal number of generators is $\le 1+c-\mid T \mid$ can be realized as the Galois group over $F$ of an extension unramified outside $p$; (ii) the OP question has a negative answer, by the same type of argument as in 1).
NB: The conclusion cannot be extended to extensions which are not $p$-extensions (for a given $p$).One could as previously introduce the maximal extension $E_S$ of $F$ which is unramified outside $S$, but this stops short here. It is not even known whether $Gal(E_S/F)$ is finitely generated.
It's not possible to do this using no information about $\mathbb{Q}$ other than that it's a field, because there are fields $F$ for which the collection of Galois groups of their finite Galois extensions is not closed under taking products. The simplest example is any finite field; here the Galois groups of finite Galois extensions are precisely the finite cyclic groups $C_n$, and e.g. $C_2 \times C_2$ is not cyclic. But any quasi-finite field will work, some of which have characteristic $0$, so even that assumption won't save you.
More abstractly, the Galois groups of finite Galois extensions of $F$ are precisely the finite quotients of the absolute Galois group $\text{Gal}(\overline{F}/F)$. The class of profinite groups which arise in this way is quite diverse, and there's no reason to expect the collection of finite quotients of some profinite group (which for finite fields is the profinite integers $\widehat{\mathbb{Z}}$) to be closed under products.