A small loop on a surface crossing a knot transversally has Gauss linking integral $\pm1$ with the knot

Let $E:\mathbb{R}^2\to\mathbb{R}^4$ be $C^2$ on a ball $B(z,r)$ with $|E|=1$ there, so $E$ is a small surface in $S^3$. Let $\gamma:\mathbb{R}\to S^3$ be a $C^2$ knot (period $1$, injective modulo $1$) which meets $E(B(z,r))$ only at $E(z)=\gamma(t_0)$, transversally:

(rows of a $4\times4$ matrix). Let $N\in S^3$ miss $\gamma$ and $E(B(z,r))$. Then for all sufficiently small $\rho>0$ the loop $s\mapsto E(z+\rho(\cos2\pi s,\sin2\pi s))$ has Gauss linking integral (gaussLinkingIntegral, stereographic projection from $N$) with $\gamma$ equal to $\operatorname{sign}(d)=\pm1$.

Suggested proof. Shift the parameter of $\gamma$ so $t_0=0$. Taylor expansion gives $E(z+\rho w)=\gamma(0)+\rho\,dE(z)w+O(\rho^2)$, while a tube estimate (the knot is embedded and $dE(z)w$ is transverse to $\gamma'(0)$ inside $T_{\gamma(0)}S^3$) keeps both $E(z+\rho w)$ and $\nu(\gamma(0)+\rho\,dE(z)w)$, $|w|=1$, at distance $\ge c\rho$ from $\gamma$, so the straight-line homotopy between them (normalized to $S^3$) misses $\gamma$ for small $\rho$. By homotopy invariance the Gauss integral equals that of the round meridian $\nu(\gamma(0)+\rho(\cos 2\pi s\,f_1+\sin2\pi s\,f_2))$, $f_i=\partial_iE(z)$, which is $1$ when $d>0$ (gaussLinkingIntegral_meridian_eq_one, using the symmetry of the Gauss integrand in the two loops). If $d<0$, reverse the meridian ($s\mapsto -s$), which changes the sign.