The answer is "no", in general, since there may be local obstructions. Suppose, for example, that $k$ and $n$ are odd prime powers, and let $L/\mathbb{Q}$ be the unique intermediate quadratic in $F$. A necessary condition for the existence of $K$ is that the cyclic degree $k$ extension $F/L$ embeds into a cyclic degree $n$ extension $K/L$. Every prime $\mathfrak{p}$ of $L$ that is totally ramified in $F$ would have to be totally ramified in $K$, so if the residue characteristic of $\mathfrak{p}$ is coprime to $n$, then you need $n$ to divide the order of the multiplicative group of the residue field of $\mathfrak{p}$. This is a genuine restriction. For concreteness, take $k=3$, $n=9$. Then there are infinitely many $D_3$ extensions of $\mathbb{Q}$ in which $7$ is split in the quadratic and ramified in the cubic, but none of them embed inside a $D_{9}$ extension, because $7$ cannot be totally tamely ramified with degree $9$.

You can upgrade such local conditions to also ensure that the bottom cyclic group of order $2$ normalises the top group and acts on it by $-1$, so that the whole extension is dihedral. If all such local conditions are satisfied, then you can prove with class field theory that the sought-for embedding always exists. In particular if $k$ is coprime with $n/k$, then the embedding will always exist. See https://arxiv.org/abs/0805.1231, section 3.1 for some examples of how such constructions work.