This answer uses the convention that $[g,h] = g^{-1}h^{-1}gh$. If $g$ is a permutation, the support of $g$, denoted $\mathrm{supp}(g)$, is the set of points that $g$ moves.
Proposition. Let $g$ and $h$ be permutations, and let $S=\mathrm{supp}(g)\cap \mathrm{supp}(h)$. Then
$$
\mathrm{supp}\bigl([g,h]\bigr) \subseteq S\cup g^{-1}(S)\cup h^{-1}(S).
$$
Proof: Suppose to the contrary that $p\in \mathrm{supp}\bigl([g,h]\bigr)$ but $p\notin S\cup g^{-1}(S)\cup h^{-1}(S)$. Since $p\notin S$, either $g$ or $h$ must fix $p$.
If $g(p)=p$, then $h$ must move $p$ since otherwise $[g,h]$ would fix $p$. Note then that $h$ moves $h(p)$ as well. Since $p\notin h^{-1}(S)$, we know that $h(p)\notin S$ and therefore $g$ must fix $h(p)$. Then $gh(p)=h(p)$, so
$
g^{-1}h^{-1}gh(p)=g^{-1}h^{-1}h(p) = g^{-1}(p)=p
$,
a contradiction.
If $h(p)=p$, then $g$ must move $p$ since otherwise $[g,h]$ would fix $p$. Note then that $g$ moves $g(p)$ as well. But since $p\notin g^{-1}(S)$ we know that $g(p)\notin S$ and therefore $h$ must fix $g(p)$. Then $h^{-1}$ fixes $g(p)$ as well, so $h^{-1}gh(p) = h^{-1}g(p)=g(p)$ and hence $g^{-1}h^{-1}gh(p) = g^{-1}g(p)=p$, a contradiction.$\quad \square$
Corollary. If $g$ and $h$ are permutations and $|\mathrm{supp}(g)\cap \mathrm{supp}(h)| = n$, then $\bigl|\mathrm{supp}\bigl([g,h]\bigr)\bigr|\leq 3n$.
This leaves open the question of which pairs $(n,k)$ are possible, where
$$
n=|\mathrm{supp}(g)\cap \mathrm{supp}(h)|\qquad\text{and}\qquad
k=\bigl|\mathrm{supp}\bigl([g,h]\bigr)\bigr|.
$$
The proposition above shows that $k\leq 3n$, and for $n=1$ we know that $(1,3)$ is the only possibility. Here are some other observations:
$k=1$ is never possible, since a nontrivial permutation must move at least two points.
$k=2$ is never possible since $[g,h]$ is always an even permutation, and the only permutations that move exactly two points are transpositions.
For $n\geq 2$, the pair $(n,0)$ is possible. For example, $g$ could be an $n$-cycle and $h$ could be the same as $g$ (or the inverse of $g$, if you want $g$ and $h$ to be distinct).
The pair $(2,3)$ is possible, e.g. $\bigl[(1\;\;2),(1\;\;2\;\;3)\bigr] = (1\;\;3\;\;2)$.
The pair $(2,4)$ is possible, e.g. $\bigl[(1\;\;2\;\;3),(1\;\;2\;\;4)\bigr] = (1\;\;2)(3\;\;4)$.
The pair $(2,5)$ is possible, e.g. $\bigl[(1\;\;2\;\;3),(1\;\;4)(2\;\;5)\bigr] = (1\;\;4\;\;5\;\;3\;\;2)$.
The pair $(3,4)$ is possible, e.g. $\bigl[(1\;\;2\;\;3),(1\;\;2)(3\;\;4)\bigr] = (1\;\;3)(2\;\;4)$.
The pair $(3,5)$ is possible, e.g. $\bigl[(1\;\;2\;\;3),(1\;\;2\;\;5)(3\;\;4)\bigr] = (1\;\;3\;\;5\;\;4\;\;2)$.
If $(n,k)$ is possible and $(n',k')$ is possible, then $(n+n',k+k')$ is possible. Specifically, if $f$ and $g$ realize $(n,k)$ and $f'$ and $g'$ realize $(n',k')$, where $\mathrm{supp}(f')\cup \mathrm{supp}(g')$ is disjoint from $\mathrm{supp}(f)\cup \mathrm{supp}(g)$, then $ff'$ and $gg'$ realize $(n+n',k+k')$.
It follows from these obervations that all pairs $(n,k)$ are possible for $n\geq 2$ and $0\leq k\leq 3n$ as long as $k\ne 1,2$.
