EDIT: I've added a proof of the claim.
Suppose that $X_1$, $X_2$ and $X_3$ intersect transversely to begin with.
Claim. You can decrease the number of intersection between $X_1$ and $X_2$ if and only if there's an embedded disc $D$ on the surface, with $\partial D \subset X_1\cup X_2$, and such that the interior of $D$ is disjoint from $X_1\cup X_2$.
When you have a third curve $X_3$, there are two possibilities: either there's a smaller disc $D'\subset D$ that eliminates an intersection of $X_3$ and $X_1$ (or $X_2$), or, if you travel along $X_3$, you enter $D$ from the $X_1$-part of the boundary and you exit from the $X_2$-part (or vice-versa). You can proceed by eliminating all the first instances (by homotoping $X_3$ across $D'$ at each step, so that $X_1$ and $X_2$ don't move: these move only decrease the number of intersection points on $X_3$), until there are no more discs like $D'$ inside $D$. Now you can just ignore $X_3$, and you can move $X_1$ across $D$, thus decreasing the number of intersection points of $X_1$ and $X_2$ (and not increasing the other intersections): Jordan curve theorem is used in some form here, since you're basically using that $D\setminus X_3$ is a union of discs.
You can do this all the way to $Y_1$ and $Y_2$ that intersect transversely and minimise the number of intersections for $X_1$ and $X_2$. You also have a curve $Y_3$ that's homotopic to $X_3$, and has fewer intersections with $Y_1$ and $Y_2$ than $X_3$ had with $X_1$ and $X_2$. You can now proceed to isotope $Y_3$ across discs $D$ as before (using $Y_1$ first, and then $Y_2$): notice that you can't have any $D'$-case here, since $Y_1$ and $Y_2$ already minimise the number of intersections.
Proof (of the claim). Take an isotopy of $X_1$ that decreases the number of intersections with $X_2$, and perturb it to obtain a generic $F:C\to\Sigma$, where $C$ is $S^1\times I$. $F^{-1}(X_2)$ is a 1-dimensional submanifold in $C$, and since $F$ decreases the intersections, there's a properly embedded arc $\gamma$ with endpoints in $S^1\times{0}$. This bounds a disc $D_0$ together with an arc in $S^1\times {0}$, and you can choose $\gamma$ so that $D_0$ is minimal with respect to inclusion.
Now also $F^{-1}(X_1)$ is a 1-dimensional submanifold of $C$ (perhaps you need to be careful with $F$ at small times, but that's easy to achieve), and if you're careful for small times, it's just a bunch of closed curves. $F^{-1}(X_1)\cap D_0$ is a bunch of closed curves and properly embedded arcs. Closed curves in $D_0$ are not "essential", in the sense that you can replace $F$ with an isotopy that doesn't have any (these curves bound discs, and you can just quotient a compact neighbourhood of these discs and take the quotient map), so we'll suppose that there are just arcs.
In $D_0$ there's a smallest disc $D_1$ (but possibly more than one) that has a boundary component along $F^{-1}(X_1)$ and one along $F^{-1}(X_2)$. $D_1$ is the preimage under $F$ of a connected component $D$ of $\Sigma\setminus(X_1\cup X_2)$, and the boundary of $D$ is nullhomotopic in $D$ (using $F|_{D_1}$), therefore $D$ is a disc.
