Let us consider the case $n=2$. Assume $p_k$ is at the origin and $p_0$ is at the point $(1,0)$ on the $x$-axis, and that $r<1$.
Distribution of $h_1$
To find the distribution of $h_1$, byBy subtracting a constant it suffices to find the distribution of $d:=d(p_1,0)$.
The density for $d$, $f_d(R)$, is proportional to $R\Theta$ where $\Theta$ is length of the interval of angles $\theta$ for which the point $(R\cos\theta,R\sin\theta)$ is within $r$ of $(1,0)$. Now, calculation shows that $$ |(R\cos\theta,R\sin\theta)-(1,0)| < r $$ is equivalent to $$ |\theta| < \arccos\left(\frac{R^2+1-r^2}{2R}\right) $$ so $f_d(R)$ is proportional to $$ 2R\arccos\left(\frac{R^2+1-r^2}{2R}\right),\quad R\in [1-r,1+r]. $$ and $f_d(R)=0$ for $R\not\in [1-r,1+r]$.
Edit: For geometric reasons (and one could also check it analytically) the integral of the given expression over the given interval is $\pi r^2$. So actually
$$ f(R)=\frac{2R\arccos\left(\frac{R^2+1-r^2}{2R}\right)}{\pi r^2},\quad 1-r\le R\le 1+r. $$
Distribution of $h_2$
The squared distance from $(1,0)+r(\cos\theta,\sin\theta)$ to the origin is $$ f(R)=\frac{2R\arccos\left(\frac{R^2+1-r^2}{2R}\right)}{\pi r^2},\quad 1-r\le R\le 1+r. $$$$ D^2 : = (r\cos\theta+1)^2+(r\sin\theta)^2 = r^2 + 1 + 2r\cos\theta$$ and the probability of picking a point $R$ or more away from the origin is then $1/\pi$ times $\theta_R>0$ where $\theta=\theta_R$ makes $D^2=R^2$. Namely $$ \theta_R = \arccos\left(\frac{R^2-r^2-1}{2r}\right) $$ Then the density is
$$ f(R)= \frac{-1}{\pi} \frac{d}{dR} \arccos\left(\frac{R^2-r^2-1}{2r}\right); $$
$$ f(R)=\frac1{\pi}\frac{1}{\sqrt{1-u(R)^2}} \frac{R}{r},\quad 1-r\le R\le 1+r, $$ where $u(R)=\frac{R^2-r^2-1}{2r}$.