There are many commensurated subgroups in a free groups which are not of the kind you described above.

We all know that normal subgroups $N\lhd \Gamma$ could be thought of as kernels of surjective maps $\Gamma\to G$. Here is an analogous way of thinking of commensurated subgroups $N<\Gamma$.

Fix a group $\Gamma$.
For every totally disconnected locally compact (tdlc) group $G$, a compact open subgroup $K<G$ and
a homomorphism $\rho:\Gamma\to G$,
$N=\rho^{-1}(K)$ is commensurable in $\Gamma$ (as $K$ is commensurtaed in $G$).
In fact, every commensurated $N<\Gamma$ is obtained this way (a construction of a group $G$ and a map $\rho$ as above is called sometimes the "Schlichting completion" of $\Gamma$ wrt $N$).

If $\rho(\Gamma)$ is dense in $G$ then (a finite index subgroup of) $K$ is normal in (a finite index subgroup of) $G$ iff (a finite index subgroup of) $N$ is normal in (a finite index subgroup of) $\Gamma$. This is easy to see.

Now, to give an example answering the question just fix a tdlc group that you know well enough (say $\text{PGL}_2(\mathbb{Q}_p)$) and a compact open subgroup which is far from being normal (say $\text{PGL}_2(\mathbb{Z}_p)$) and find an injective dense homomorphism $\rho:F_n\to G$ by fixing any generic enough $n$-tuple in $G$ (which could be made very concrete).