The inward normal coordinate is intrinsic distance from the boundary

Let $J$ be a variation field obtained by varying $p$ along $S$ while following the unit normal geodesics. Write $U={\mathrm{\partial }}_{s}F$. Then $h(U,U)=1$ and ${\mathrm{\nabla }}_{U}U=0$. Torsion freeness for this variation and metric compatibility give ${\mathrm{\partial }}_{s}h(J,U)=h({\mathrm{\nabla }}_{U}J,U)=h({\mathrm{\nabla }}_{J}U,U)=\frac{1}{2}Jh(U,U)=0$. At $s=0$, $J\in TS$ and $U=\nu$, so this inner product is zero. Thus the metric has no mixed $ds$ terms and its $d{s}^2$ coefficient is one: $F^{\ast }h=d{s}^2+h_s$. Smoothness and positivity of $h_s$ follow from the collar embedding.

In the infimum defining $\rho (x)$, boundary points in other components contribute only $+\mathrm{\infty }$; all actual paths from $S$ to $x$ lie in the component of $x$. Let the larger available collar have width $a$ and take $x=F(p,s)$ with $s<a\mathrm{/}4$. Its radial normal path has length $s$, giving $\rho (x)\le s$. Every path from $S$ to $x$ staying in the width-$a\mathrm{/}2$ collar has length at least $\int \mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }\ge s$ by the Gaussian metric form. A path leaving that collar must first meet its outer face $s=a\mathrm{/}2$, since there is no other interior boundary. Its initial portion already has length at least $a\mathrm{/}2>s$. Hence allowing paths outside the collar cannot shorten the distance. Taking the infimum gives $\rho (x)=s$.

In these coordinates $d\rho =ds$ and the Gaussian metric dual is ${\mathrm{\partial }}_s$. Therefore $\mathrm{\nabla }\rho ={\mathrm{\partial }}_s$ and its norm is one. Both extend smoothly to the boundary with inward direction. This proves all conclusions on the smaller collar, with no cut-locus assertion outside it.