I'm a bit late but here are a few remarks on Kevin's example :
There is a unique $\mathfrak{A}_4$-extension of $\mathbf{Q}_2$ because there is a unique cyclic cubic extension $C$ (namely the unramified one), the group $G=\mathrm{Gal}(C|\mathbf{Q}_2)=\mathbf{Z}/3\mathbf{Z}$ has a unique irreducible degree-$2$ $\mathbf{F}_2$-representation $\rho$, and $\rho$ occurs with multiplicity $1$ in $C^\times/C^{\times2}$. If $G$ acts on $D\subset C^\times/C^{\times2}$ through $\rho$, then Kevin's $M$ is $M=C(\sqrt D)$.
Kevin's $K$ is not galoisian over $\mathbf{Q}_2$ so there is no quadratic extension $L$ of $\mathbf{Q}_2$ such that $K=LC$.
You can find $\mathfrak{A}_4$-extensions of ramification index $4$ and residual degree $3$ over every local field $F$ with finite residue field of characteristic $2$ (although they might not be unique), so the argument can be made to work over every such $F$.
$\mathfrak{A}_4$-extensions are galoisian closures of primitive quartic extensions. I'm confident that the same trick can be applied over local fields with finite residue of characteristic $p$ (arbitrary prime) by working with primitive extensions of degree $p^2$. How does one find such extensions ? See https://arxiv.org/abs/1608.04183.