What we want to show is
$$\begin{aligned}&[K_{1}K_{2} : F] \leq \frac{[K_{1} : F][K_{2} : F]}{[K_{1} \cap K_{2} : F]} \\\iff &[K_{1}K_{2} : K_{1} \cap K_{2}][K_{1} \cap K_{2} : F] \leq [K_{1} : F][K_{2} : K_{1} \cap K_{2}] \\\iff & \frac{[K_{1}K_{2} : K_{1} \cap K_{2}]}{[K_{2} : K_{1} \cap K_{2}]} \leq \frac{[K_{1} : F]}{[K_{1} \cap K_{2} : F]} \\\iff & [K_{1} K_{2} : K_{2}] \leq [K_{1} : K_{1} \cap K_{2}].\end{aligned}$$
Let $E = K_{1} \cap K_{2}$. Suppose $[K_{1} : E] = n$ and $K_{1} = E\alpha_{1} + \cdots + E\alpha_{n}$ where $\alpha_{1}, \dots, \alpha_{n} \in K_{1}$. Then $K_{1} = E(\alpha_{1}, \dots, \alpha_{n}) \implies K_{1}K_{2} = K_{2}(\alpha_{1}, \dots, \alpha_{n})$. Now, $[K_{1} : E] < \infty$ implies that $\alpha_{1}, \dots, \alpha_{n}$ are algebraic over $E$, hence algebraic over $K_{2}$. Then we have $K_{1}K_{2} = K_{2}[\alpha_{1}, \dots, \alpha_{n}]$.
To show that $[K_{1}K_{2} : K_{2}] < n$, it suffices to show that $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. Now every element of $K_{1}K_{2}$ is of the form of linear combinations of $K_{2}\alpha_{1}^{r_{1}}\cdots\alpha_{n}^{r_{n}}$.Since each $\alpha_{i} \in K_{1}$, their multiple is also in $K_{1}$, i.e. in $E\alpha_{1} + \dots + E\alpha_{n} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. So is the combinations, i.e. the element of $K_{1}K_{2}$. Hence we have shown $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$.