The positive-push Seifert pairing is well-defined for a link surface

Fix a finite cellulation of $F$ from [F1]. An integral cellular $1$-cycle represents each class of $H_1(F;\mathbb{Z})$. Replace each edge with coefficient $n$ by $\mathrm{\mid }n\mathrm{\mid }$ parallel oriented copies, reversing orientation when $n<0$. At each vertex the cycle condition says that incoming and outgoing copies occur in equal number. Around the boundary of a small vertex disk there is therefore an adjacent incoming/outgoing pair; join that pair by a small arc and delete its endpoints, repeating finitely until all endpoints are paired. The resulting arcs can be chosen disjoint in the disk and oriented consistently with the copies. Each smoothing changes the chain only by the boundary of a $2$-chain in that disk. The result is a finite disjoint union of oriented embedded circles representing the original homology class.

For representatives $X,X^{\mathrm{\prime }}$ of the same class in $H_1(F;\mathbb{Z})$, there is an integral singular $2$-chain $A$ in $F$ with $\mathrm{\partial }A=X-X^{\mathrm{\prime }}$. Push the chain through a small positive normal level of $F$; its boundary is $X^{+}-X^{\mathrm{\prime }+}$ and it is disjoint from any representative $Y\subset F$. Thus $X^{+}$ and $X^{\mathrm{\prime }+}$ represent the same class in $H_1(S^3\setminus Y;\mathbb{Z})$, so their meridian-coordinate linking sums with $Y$ agree. Varying the small positive level traces an annulus in the positive bicollar, also disjoint from $Y$, and gives the same class.

If $Y,Y^{\mathrm{\prime }}$ represent the same class, choose an integral $2$-chain $B$ in $F$ with $\mathrm{\partial }B=Y-Y^{\mathrm{\prime }}$. For each circle component of $X^{+}$ and each circle component of $Y-Y^{\mathrm{\prime }}$, [F4] supplies a spanning surface and [F5] permits a homotopy of the other circle in its complement to make it transverse to that surface while remaining an embedded circle. By [F3] choose a regular diagram of the pair. By [F2], linking is the half-sum of mixed crossing signs; exchanging the two components leaves the same mixed crossings with the same signs, so linking is symmetric. Bilinearity then gives $\mathrm{lk}(X^{+},Y-Y^{\mathrm{\prime }})=\mathrm{lk}(Y-Y^{\mathrm{\prime }},X^{+})$. The cycle $Y-Y^{\mathrm{\prime }}$ bounds $B$ in $S^3\setminus X^{+}$, so its total meridian coefficient there is zero. Hence the two linking sums agree in the second variable as well. Additivity of meridian coefficients gives bilinearity in both variables.

The value in the Statement is therefore independent of the chosen cycle representatives and push-off level and is bilinear on homology. In any ordered integral basis its matrix entries are $V_{ij}=\mathrm{lk}(a_i^{+},a_j)$, as required.