$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$In general, it is not true that $Gal(L/\QQ) \cong Gal(K/\QQ) \ltimes \mathrm{Cl}(K)$.
What is true
Suppose that we have a short exact sequence of groups
$$0 \to A \to H \to G \to 0$$
where $A$ is abelian. Since $A$ is normal, the group $H$ acts on $A$ by conjugation; since $A$ is abelian, this action factors through $H/A \cong G$. So, in any such setting, we get an action of $G$ on $A$.
In your notation, Galois theory gives us
$$0 \to \mathrm{Cl}(K) \cong Gal(L/K) \to Gal(L/\QQ) \to Gal(K/\QQ) \to 0.$$
It turns out that this action of $Gal(K/\QQ)$ is the same as the natural action of $Gal(K/\QQ)$ on the class group. I don't know a reference for this, but it's not hard to see that $\sigma \cdot \mathrm{Frob}_{\mathfrak{p}} \cdot \sigma^{-1} = \mathrm{Frob}_{\sigma \mathfrak{p}}$ for an unramified prime $\mathfrak{p}$ of $K$ and an element $\sigma \in G$. The result follows from this plus an extremely weak version of Cebataorv density (just enough to know that the values of Frobenius generate the Galois group).
However, this does not mean that the short exact sequence is semidirect and, as we will see below, it need not be.
A strategy for finding a counterexample
Suppose that we could construct a Galois extension $F/\QQ$, with Galois group $H$, and an abelian subgroup $A$ of $H$ such that
(1) The sequence $0 \to A \to H \to H/A \to 0$ did not split and
(2) For every prime $\mathfrak{p}$ of $F$ (including the infinite primes), the inertia group $I_{\mathfrak{p}}$ is disjoint from $A$.
I claim that taking $K$ to be the fixed field of $A$ will give a counterexample.
The extension $F/K$ is abelian (since $A$ is abelian) and unramified (by the condition on inertia groups). So we have $F \subseteq L$ and we have a commuting diagram
$$\begin{matrix}
0 &\to& \mathrm{Cl}(K) &\longrightarrow& Gal(L/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
& & \downarrow & & \downarrow & & \| & & \\
0 &\to& A &\longrightarrow& Gal(F/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
\end{matrix}$$
Suppose for the sake of contradiction that we had a map $Gal(K/\QQ) \to Gal(L/\QQ)$ splitting the top sequence. Then the composite map $Gal(K/\QQ) \to Gal(L/\QQ) \to Gal(F/\QQ)$ would split the bottom sequence, contradicting (1).
This makes it clear why counterexamples are hard to find. Since $\QQ$ has no unramified extensions, one of the $I_\mathfrak{p}$ must be nontrivial. So, just on the level of group theory, we need to find a non-semidirect short exact sequence $0 \to A \to H \to G \to 0$, with $A$ abelian and a nontrivial subgroup $I \subset H$ so that $I \cap A = \{ e \}$. If $I \to G$ is surjective, this implies that the sequence is semidirect after all. Just on the group theory level, we see that there are no examples with $G$ a cyclic group of prime order.
A counterexample
Let $\zeta$ be a primitive $85$-th root of unity. Recall that $$Gal(\QQ(\zeta)/\QQ) \cong (\ZZ/85)^{\ast} \cong (\ZZ/5)^{\ast} \times (\ZZ/17)^{\ast} \cong (\ZZ/4) \times (\ZZ/16).$$
We will always write elements of this group as ordered pairs in $(\ZZ/4) \times (\ZZ/16)$.
Let $F$ be the fixed field of $\{ 0 \} \times (4 \ZZ/16)$. So $Gal(F/\QQ) \cong (\ZZ/4) \times (\ZZ/4)$. The only ramified primes are $5$, $17$ and $\infty$ with inertia groups $\ZZ/4 \times \{0 \}$, $\{ 0 \} \times \ZZ/4$ and $(2,0)$ respectively. Let $A$ be the order $2$ subgroup generated by $(2,2)$. We see that $A \cap I_{\mathfrak{p}}$ is trivial for every $\mathfrak{p}$. Finally, we have $A \cong \ZZ/2$, $H \cong (\ZZ/4) \times (\ZZ/4)$ and $H/A \cong (\ZZ/4) \times (\ZZ/2)$, so the sequence is not semidirect.
I have not actually computed $\mathrm{Cl}(K)$ for this example.
One can build similar counterexamples whenever $H \cong (\ZZ/q^2)^2$ for some prime $q$, being a little careful about the prime at $\infty$ if $q=2$.
Smaller counterexample
I messed around a bit more, and concluded that there is nothing stopping me from taking $H$ to be the dihedral group of order $8$ and $A$ to be its (two-element) center. For example, let $F$ be the splitting field of $\QQ(\sqrt{1+8 \sqrt{-3}})$ (so $K = \QQ(\sqrt{-3}, \sqrt{193})$). If I haven't made any dumb errors, the only ramified primes are $3$, $193$ and $\infty$, and the inertia groups are all noncentral two-element subgroups. Even if I got this particular example wrong, I'm pretty sure there is no general obstacle to making an example like this.
"Your" notations are a bit confusing, but let us keep them, just adding $m'=m/p^{e_p}$, so that $p \nmid m'$. Then known classical results on cyclotomic extensions assert that $\mathbf Q_p (\zeta_m)/\mathbf Q_p$ is the compositum of the two linearly disjoint extensions $\mathbf Q_p (\zeta_{m'})$ and $\mathbf Q_p (\zeta_{p^{e_p}})$. In the same way,
$L_\mathfrak q=K_\mathfrak p(\zeta_m)$ is the compositum of $\mathbf Q_p (\zeta_{p^{e_p}})$ and $\mathbf Q_p (\zeta_{m''})$ (containing $\mathbf Q_p (\zeta_{m'})$), with $p\nmid m''$. Concerning ramification, the branch $\mathbf Q_p (\zeta_{p^{e_p}})/\mathbf Q_p$ is totally ramified, whereas the branches $\mathbf Q_p (\zeta_{m''})/\mathbf Q_p$ and $L_\mathfrak q/\mathbf Q_p (\zeta_{p^{e_p}})$ are unramified, so that $Gal(L_\mathfrak q /\mathbf Q_p (\zeta_{m''}))\cong Gal(\mathbf Q_p (\zeta_m)/\mathbf Q_p (\zeta_{m'}))\cong Gal(\mathbf Q_p (\zeta_{p^{e_p}})/\mathbf Q_p)$ and the corresponding extensions are totally ramified.
By the above, the inertia subfield $F$ (bad notation, there should be an index indicating that this is a local field) coincides with $\mathbf Q_p (\zeta_{m''})$, and $L_\mathfrak q=F(\zeta_m)=F(\zeta_{p^{e_p}})$ . NB: things get clearer with a Galois diagram.
Best Answer
I don't know if this is helpful, but for $E = \mathbb{Q}(\sqrt{-6})$ case....
You know that $H/E$ is quadratic, and thus $H = E(\sqrt{D})$ for some $D \in E$. $H/E$ is an unramified extension, however all number fields are ramified over $\mathbb{Q}$, so clearly $D$ must factor into ramified primes of $E$.
I looked over $2$ first. I don't remember why, but I convinced myself at one time that for a quadratic extension to be unramified, you need to find a $D$ whose prime factorization is square. (hopefully this is fresh in your mind so you'll be able to work out why!) So I observe
$$(2, \sqrt{-6})^2 = (2)$$
However, $2$ is clearly (?) not a square. Thus $E(\sqrt{2})$ is an unramified extension.
I also vaguely recall a general theorem that the Hilbert class field of $\mathbb{Q}(\sqrt{m})$ consists of adjoining square roots of the prime factors of $m$, with appropriate signs.