I'm a bit late but here are a few remarks on Kevin's example :

1.  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 $M=C(\sqrt D)$.

2.  Kevin's $K$ is not galoisian over $\mathbf{Q}_2$ and nor is any of its unramified extensions, so no  such $L$ exists. 

3.  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 $F$.

4. $\mathfrak{A}_4$-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.