There is a section in an explanation about knot surgery which I do not understand in "Knot surgery revisited" by Fintushel, p.203.
Let $X$ be a simply-connected compact $4$-manifold. Let $K$ be a knot in $S^3$.
Let $M_K$ be the $3$-manifold obtained by performing $0$-framed surgery on $K$. After this surgery, the longitude of $K$ spans a disc (before it, the meridian $m$ spanned a disc). The meridian $m$ is as well an embedded circle in $M_K$.
Fintushel writes: "In $S^1\times M_K$ we have the torus $T_m:=S^1\times m$ of self-intersection $0$." I suppose this means that if let $a\in H_2(X;\mathbb{Z})$ is the homology of the surface $T_m$, then $Q_X(a,a)=0$.
How does one show $Q_X(a,a)=0$ for the intersection form $Q_X:H^2(X,\partial X;\mathrm{Z})\times Q_X:H^2(X,\partial X;\mathrm{Z})\to \mathbb{Z}$?
One idea: Let $F_1$ and $F_2$ be surfaces that represent $a$. I thought about using the description of $Q_X(a,a)$ as sum of intersection points counted with sign, but I do not know how to do that.
Take any embedded annulus $A : S^1 \times [0,1] \to M_K$ such that $A(S^1 \times 0)=m$. Since $A$ is embedded, $m'=A(S^1 \times 1)$ is disjoint from $m$. We may regard $A$ as a homotopy in $M_K$ from $m$ to $m'$, and crossing with $S^1$ we get a homotopy in $S^1 \times M_K$ from $F_1 = T_m = S^1 \times m$ to the disjoint torus $F_2 = T_{m'} = S^1 \times m'$. The intersection number of a pair of homology classes represented by disjoint oriented submanifolds equals zero. So the self-intersection number of a homology class represented by a disjoint pair of oriented submanifolds equals zero.
So basically your idea is right and is very simple if you notice that $F_1,F_2$ can be chosen to be disjoint.