We say a collection of closed curves $\{\alpha_1,...,\alpha_k\}$ on a surface $S_{g,n}$ (a surface of genus $g$ with $n$ boundary components) fill the surface $S_{g,n}$, if $S_{g,n}-\{\alpha_1,...,\alpha_k\}$ is a collection of disks and annuli.
Let us call the red curve $A$ and the blue curve $B$ in the following picture. It is clear that $\{A,B\}$ fill the surface $S_{0,4}$ (a disk with 3 holes) in the figure. For an element $\phi\in MCG(S_{0,4})$(=mapping class group of a disk with 3 holes), let $A'=\phi(A)$ and $B'=\phi(B)$. Is it true that $\{A',B'\}$ fill the surface $S_{0,4}$ as well?
Yes!
We can prove this using the following fact lifted from A Primer on Mapping Class Groups by Farb and Margalit:
Since your map $\phi$ is orientation preserving, the surface obtained by cutting along $A$ versus along $\phi(A)$ are homeomorphic for each simple closed curve in your collection. Therefore, in general if $\{\alpha_i\}$ cuts your surface into disks and annuli, then $\{\phi(\alpha_i)\}$ will cut your surface into exactly the same number of disks and annuli.