The push-off of the boundary of an embedded disk along its $\xi$-projected conormal has linking number $0$

Let $S=H^{-1}(1)$ be strictly star-shaped, $P=(x,T)$ a prime periodic orbit, and $e$ a smooth embedding of the closed unit disk into $S$ with $e(S^1)=x(\mathbb{R})$, and $u(s)$ a smooth $1$-periodic lift with $e(u(s))=x(Ts)$. Push $P$ off along the $\xi$-projection of the outward conormal $n(s)=de(u(s))\,u(s)$,

Then for all small $\varepsilon>0$ the loops $s\mapsto x(Ts)/|x(Ts)|$ and $\gamma_\varepsilon$ in $S^3$ have linking number $0$, in the sense of IsLinkingNumber (Gauss integral after stereographic projection).

Proof idea: $\pi_\xi n$ and $n$ differ by a multiple of $X_H$, which is tangent to $P$, so the two push-offs are homotopic in the complement of $P$. The push-off along $n$ is close to $e((1+\varepsilon)u(s))$, which bounds the disk $\rho\mapsto e(\rho\,u)$, $\rho\le 1+\varepsilon$, missing $P$ after radial projection. So it is null-homotopic in $S^3\setminus P$, and the Gauss integral vanishes. This is the statement that the Seifert framing of a knot bounding a disk is $0$.