In this setting $L/K$ will always be separable.
To see that let $z\in L\setminus K$ be arbitrary. Because $L/K$ is finite, $z$ is algebraic over $K$. Therefore $z$ has a minimal polynomial $m(x)\in K[x]$. Over $\overline{F}$ $m(x)$ splits into linear factors
$$
m(x)=(x-z_1)(x-z_2)\cdots (x-z_n)
$$
with $z_1=z$. The elements $\sigma^i(z)$, $i=0,1,2,\ldots,$ are all roots of $m(x)$, so they are among the $z_j$:s. In particular there are only finitely many $\sigma^i(z)$:s. Let $k$ be the smallest positive integer such that $\sigma^k(z)$ is among the $\sigma^j(z), 0\le j<k$. Because $\sigma$ is bijective this implies that $\sigma^k(z)=z$.
Next consider the polynomial
$$
f(x)=(x-z)(x-\sigma(z))(x-\sigma^2(z))\cdots (x-\sigma^{k-1}(z)).
$$
Its zeros are all distinct. And we see that the coefficients of $f$ are fixed under $\sigma$. Therefore $f(x)\in K[x]$. This implies that $m(x)$, as the minimal polynomial, is a factor of $f(x)$. That shows that $m(x)$ is separable proving the claim.
We also see that we must have $m(x)=f(x)$. I guess that we could turn that into a solution by-passing your separability concern altogether, but I haven't thought it through.
