A knot crossing a disk transversally, always from the same side and at least once, has non-zero Gauss linking number with the disk's boundary

Let $E:\mathbb{R}^2\to\mathbb{R}^4$ be $C^2$ on an open neighbourhood of the closed unit disk $\overline{\mathbb D}$ with $|E|=1$ on $\overline{\mathbb D}$, so $E(\overline{\mathbb D})\subset S^3$. Let $u:\mathbb{R}\to S^1$ be a $C^2$, $1$-periodic immersion ($u'\neq0$), so $s\mapsto E(u(s))$ runs (possibly several times) around $E(S^1)$. Let $\gamma:\mathbb{R}\to S^3$ be a $C^2$, $1$-periodic loop that is injective modulo $1$ (a knot). Assume:

1. $\gamma$ misses the boundary loop: $E(u(s))\neq\gamma(t)$ for all $s,t$;
2. $\gamma$ crosses the disk transversally, always from the same side: for a fixed $\sigma\in\mathbb{R}$,

3. $\gamma$ meets $E(\overline{\mathbb D})$ at least once.

Then there is a unit vector $N$ missing both loops such that the Gauss linking integral gaussLinkingIntegral N (E ∘ u) γ (stereographic projection from $N$) is a non-zero integer.

This is the statement "the linking number of a knot with the boundary of a surface equals the algebraic intersection number of the knot with the surface", in the special case where all intersections have the same sign. The algebraic count is then $\pm\deg(u)\cdot\#\{t\in[0,1):\gamma(t)\in E(\overline{\mathbb D})\}\neq0$. Here $E$ need not be injective, and $\det$ is the $4\times4$ determinant with the four vectors as rows, which orients the disk relative to $\gamma$ inside $S^3$.

Suggested proof. The crossings are finitely many interior points $v_j$, by transversality and compactness. Choose a centre $c$ in the open disk so that the $v_j$ lie on distinct rays from $c$; the bad centres form finitely many lines. Work in the strip coordinates $(\lambda,s)\mapsto c+\lambda(u(s)-c)$, $\lambda\in[0,1]$. Graph loops $s\mapsto E(c+h(s)(u(s)-c))$ interpolate between $h\equiv1$ (the boundary loop) and $h\equiv0$ (a constant loop, Gauss integral $0$). Use bump functions to move $h$ past one crossing at a time. Each move changes the Gauss integral by the Gauss integral of a small loop around that crossing: arcs with fixed endpoints can be homotoped freely, since the boundary terms of the variation formula vanish. A small loop on a transverse surface around a knot has Gauss integral $\pm1$, with the sign given by condition 2 and by the orientation of $u$, which is the same at every crossing.