Suppose $E_1/F$ and $E_2/F$ are finite field extensions. The degree of the composite field $E_1E_2$ over $F$ is less or equal to the product of the degree of $E_1$ over $F$ times the degree of $E_2$ over $F$, i.e. $$ [E_1E_2:F] \leq [E_1:F] [E_2:F]. $$ See for example: The degree of a field extension is smaller than the product of the degrees of field extensions of intermediate fields that generate the field.
I suspect that $[E_1E_2:F]$ divides the product $[E_1:F] [E_2:F]$. Do you have a suggestion on how to prove this? Note we have the towers $F\subset E_1\subset E_1E_2$ and $F\subset E_2\subset E_1E_2$ that might be useful.
It need not divide the product.
Consider $E_1=\mathbb{Q}(\sqrt[3]{2})$, and $E_2=\mathbb{Q}(\zeta\sqrt[3]{2})$, where $\zeta$ is a complex primitive cubic root of unity. Each of those has degree $3$ over $\mathbb{Q}$, because they are given by extending with a root of the irreducible polynomial $x^3-2$.
The compositum $E_1E_2=\mathbb{Q}(\sqrt[3]{2},\zeta)$ is the splitting field of $x^3-2$, which has degree $6$ over $\mathbb{Q}$. But $6$ does not divide $[E_1:\mathbb{Q}][E_2:\mathbb{Q}]=9$.
Of course, if $\gcd([E_1:F],[E_2:F])=1$, then the degree of the compositum will be equal to the product, since it will be a multiple of each and less than or equal to their product.