I know that when $K/F$ and $L/F$ are finite Galois extension (where $L, K<\overline{F}$), $[KL:L]$ divides $[K:F]$, and $[KL:L]=[K:F]$ iff $K\cap L=F$.
My question is that above proposition is still valid for finite separable extension $K/F$ and $L/F$. I couldn't find counterexample...
Take $F = \Bbb Q$, $L = \Bbb Q(\sqrt[3]2)$ (embedded in $\Bbb R$) and $K = \Bbb Q(\sqrt[3]2\omega)$ where $\omega = e^{\frac{2\pi i}3}$ is a primitive root of unity.
Then $KL = \Bbb Q(\sqrt[3]2, \omega)$ has degree $6$ over $\Bbb Q$ and hence $[KL : L] = 2 < [K:F]$. However it is clear that $K \cap L = F$.
To explain a bit the general situation, I have answered a MathOverFlow question on the tensor product of two fields (of characteristic zero) in which I explained that $K \otimes_F L$ is isomorphic to the direct product of all compositums of $K$ and $L$ over $F$.
Since $K \otimes_F L$ has dimension $[K:F][L:F]$ over $F$, it is clear that we have $[KL:L] = [K:F]$ iff $KL$ is the only compositum of $K$ and $L$ over $F$. This usually is not the case for non-Galois extensions.