Say we are given a disconnected closed orientable surface $S=S_1\coprod S_2$ with $f=f_1\coprod f_2:S\rightarrow M$ such that the $f_i$ are embeddings and the images are incompressible.
Suppose that $f_1(S_1)\cap f_2(S_2)=F$ is a subsurface is it possible to say isotope $f_2$ such that the images are disjoint?
I was thinking of pushing along the normal to $f_1(S_1)$, to make them disjoint.
On
Consider the situation where $S_1 \hookrightarrow M$ is incompressible. Then take any $\psi \in H^1M = Hom(H_1M ,\mathbb Z)$, such that the pullback $\psi^*$ is non-trivial, i.e. $H_1 S \to H_1M \to \mathbb Z$ is non-trivial (such elements exist if the image of $H_1S$ has $b_1>0$). Then every embedded surface $S_2\hookrightarrow M$ Poincaré dual to $\psi$ will intersect $S_1$ non-trivially.
However after surgering you can obtain multiple disjoint surfaces, homologous to $S$. After surgering again you can even get a bunch of incompressible ones.
I doubt it, and this might be an example:
Let $$ M = S^1 \times S^1 \times S^1 $$ and let $S_1 = S_2 = S^1 \times S^1$.
Embed $S_1$ as $S^1 \times S^1 \times \{p\}$ and $S_2$ as $\{q\} \times S^1 \times S^1$, where $p$ and $q$ are elements of $S^1$. I believe that you cannot move one of these off the other, and I suspect that might be true for homology/cohomology reasons, but I can't see a clear argument offhand.
But here's a rough approximation of an argument: look at the intersect of these two surfaces with $$ U = S^1 \times \{a\} \times S^1. $$ It consists of a pair of circles that intersect once (a meridian and a longitude). During a (smooth enough) deformation, the intersections $S_i \cap U$ will change "nicely" (i.e., there'll be a homology between the "before" and "after" intersections given by the intersection of the whole deformation with $U$). But in the "before" situation, the two intersections meet transversely in a single point, and in the after situation, they would be disjoint. So you'd have a homology between a single point and the empty set, and that's not possible.