Let $K\supseteq F$ be fields of characteristic $0$ such that any element of $K$ has degree $\leq n$ over $F$.
If $[K:F]=1$ we are done, and if not then there is some $r_1\in K\setminus F$. Let $E_1=F(r_1)$. Observe that $[E_1:F]\leq n$ by hypothesis. If $[K:E_1]=1$ then $[K:F]=[E_1:F]$ and we are done, and if not then there is some $r_2\in K\setminus E_1$, and let $E_2=E_1(r_2)=F(r_1,r_2)$. Note that because $E_2$ is a finite separable extension of $F$, it is simple, say with $E_2=F(s)$. Thus, again by hypothesis, we have $[E_2:F]\leq n$. It is clear that we can repeat this argument any finite number of times. If we have not found an $m$ for which $K=E_m$ by the time we reach $E_d$ where $d=\lceil\log_2(n)\rceil+1$, then
$$[E_d:F]=\underbrace{[E_d:E_{d-1}]}_{\geq 2}\cdots\underbrace{[E_1:F]}_{\geq 2}>n$$
but this contradicts $[E_d: F]\leq n$. Thus $K=E_m$ for some $m$, so that $[K:F]\leq n$.
(I've edited out my previous, unnecessarily-complicated argument using Zorn's lemma. Thanks to Mariano for catching that there was a simpler way.)
As Curtis points out, it is possible for every element of $K$ to have finite degree over $F$, while the extension $K/F$ has infinite degree. For example, take $F=\mathbb{Q}$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)$, or $K=\overline{\mathbb{Q}}$. Note that finite is different than bounded.
Edit: sorry misread the question! For instance take $\mathbf{F}_{p^2}((t))/\mathbf{F}_p((t))$ and $\mathbf{F}_p((t^{1/2}))/\mathbf{F}_p((t))$. These are both degree 2 extensions, but are not isomorphic: in particular the second one is isomorphic to $\mathbf{F}_p((t))$ itself, which is not isomorphic to $\mathbf{F}_{p^2}((t))$.
Let's show that these are degree 2 extensions. Note $\mathbf{F}_{p^2}/\mathbf{F}_p$ is a degree $2$ extension, given by adjoining a root of an irreducible polynomial $f(x) \in \mathbf{F}_p[x]$ of degree $2$ over $\mathbf{F}_p$ (for instance, if $p$ is odd, take $x^2 - r$, where $r \in \mathbf{F}_p$ is not a quadratic residue mod $p$). Then the extension $\mathbf{F}_{p^2}((t))/\mathbf{F}_p((t))$ is given by, again, adjoining a root of $f(x)$, this time viewed as a polynomial in $\mathbf{F}_p((t))[x]$: note it's still irreducible in this ring, and thus the extension is of degree $2$. On the other hand, you could adjoin a root of the polynomial $x^2 - t \in \mathbf{F}_p((t))[x]$ and then you get the extension $\mathbf{F}_p((t^{1/2}))$: this is again of degree $2$ since the polynomial has degree $2$.
More generally this illustrates the distinction between ramified and unramified extensions of equal characteristic local fields.
You could also do this starting with $\mathbf{F}_p(t)$, the field of rational functions in one variable over $\mathbf{F}_p$, instead of $\mathbf{F}_p((t))$ the field of Laurent series in one variable over $\mathbf{F}_p$, it wouldn't change anything above, and you could come up with more examples. I just chose Laurent series because the theory of finite field extensions of a local fields is a bit simpler in the end.
Best Answer
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.