The $\xi$-projected outward conormal of a disk transverse to the flow winds once relative to the global frame

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 $X_H(e(u))\notin \operatorname{im} de(u)$ for $|u|<1$. Let $u(s)$ be a smooth $1$-periodic lift with $e(u(s))=x(Ts)$. Let $n(s)=de(u(s))\,u(s)$ be the outward conormal of the disk along its boundary, and let

be its projection to $\xi=\ker\lambda_0\cap TS$ along $X_H$. Then $\pi_\xi n(s)=r(s)\big(\cos\theta(s)\,Z_1+\sin\theta(s)\,Z_2\big)$ at $x(Ts)$, with smooth $r>0$ $1$-periodic and smooth $\theta$ satisfying

That is, the conormal winds exactly once, positively, relative to the global frame $(Z_1,Z_2)$ of $\xi$, where $\omega_0(Z_1,Z_2)=1$.

Proof idea: $\pi_\xi n$ is intrinsic to the disk, so we may reflect $e$ to make $d\lambda_0|_D>0$. Transversality makes the projection along $X_H$ an orientation-preserving isomorphism $TD\to\xi$ over the open disk. The frame $(\pi_\xi\partial_1 e,\pi_\xi\partial_2 e)$ extends over the disk, so it has winding $0$ relative to $(Z_1,Z_2)$, and the radial vector $\pi_\xi\partial_r e$ winds $+1$. At the boundary, $\ker(\pi_\xi|_{TD})=\mathbb{R}X_H=T\partial D$, so $\pi_\xi\partial_r e\ne0$ up to $|u|=1$. By Stokes, the boundary orientation matches the flow direction.