Suppose that $F$ and $G$ are compact surfaces and $h:F\rightarrow G$ is a homotopy equivalence. I believe that there is a theorem that gives a standard form for homotopy equivalences, but I don't remember where it is proved. Here is what I think the theorem is.
The flip is the map $f:[0,1]\times [0,1]\rightarrow [0,1]\times [0,1] $ given by $$f(s,t)=(s-t(2s-1),t).$$ When $t=0$ this is just the identity map on the interval $[0,1]$ in the variable $s$ and when $t=1$ this is the map $f(s)=1-s$ that flips the interval.
Suppose that $\alpha$ is a properly embedded arc in $F$. Let $F'$ be the result of cutting $F$ along $\alpha$ and regluing so that one side of $\alpha$ gets glued to the other side with the direction reversed. The surface $F'$ is homotopy equivalent to $F$. The surface $F'$ will usually be of a different topological type than $F$. If the endpoints of $\alpha$ are in distinct boundary components of $F$ then $F'$ has fewer boundary components. If the endpoints of $\alpha$ are in the same boundary component of $F$ then $F'$ may have more boundary components.
The flip along $\alpha$ is a map $f_\alpha:F\rightarrow F'$ that is the identity map away from a regular neighborhood of $\alpha$ and inside that regular neighborhood is a flip, where you see the to interior arcs bounding the regular neighborhood as $t=0$ and $t=1$, and the arc $\alpha$ given by $t=\frac{1}{2}$. The flip along $\alpha$ is a homotopy equivalence.
I believe:
Theorem: If $h:F\rightarrow G$ is a homotopy equivalence of compact surfaces then there is a map $g:F\rightarrow G$ that is homotopic to $h$ that is the result of applying a finite number of flips along a disjoint system of proper arcs followed by a homeomorphism.
The idea of proof is this. We know that if $f:F\rightarrow G$ is a homotopy equivalence that sends the peripheral curves to curves that are homotopic to peripheral curves then $f$ is homotopic to a homeomorphism.
Look at the image in $G$ of the boundary curves of $F$ under $h$ , with $h$ in general position, which means there are only double points of transverse intersection of the image of the boundary components of $F$ under $h$ .
The inverse image of a double point can be seen as a proper $1$-manifold embedded in $F$. That proper $1$-manifold contains an arc joining the two points in the boundary of $F$ that are mapped to the double point, form $F'$ by doing the flip along the inverse image of the double point, and build a map from $F'$ to $G$ that is the identity away from the double point, and the image of its boundary is the result of smoothing the crossing. Continue on, until there are no double points.
I claim that a homotopy equivalence of surfaces $f:F'\rightarrow G$ that maps the boundary of $F'$ to simple closed curves has the property ( probably except in some degenerate cases) that the image of the boundary curves of the surface $F'$ under $f$ are homotopic to the boundary components of $G$.
To see this make $f$ transverse near the boundary and then deform away from the boundary so that it is self transverse. The inverse image of $f(\partial F')$ under $f$ is a closed $1$-dimensional submanifold of $F'$. If some component is not boundary parallel, then you can show $f_{\#}$ is not injective on $\pi_1(F')$, so $f$ was not a homotopy equivalence. You can now homotope $f$ to be a homeomorphism onto its image, and $f(F')$ is a subsurface of $G$ that is homeomorphic to $G$. Hence its boundary is parallel to $\partial G$.
Therefore after ``smoothing'' the self intersections of $h:F\rightarrow G$ by flips to get rid of self intersections we have a map that is homotopic to a homeomorphism of surfaces.
Maybe, I'm wrong, and that would be a good answer too. However I think this theorem if it is true is in the literature, and I don't know where.