= Solution
Here $\Lambda$ is planar <Lebesgue measure>. For a Borel set $B\subseteq[0,\infty)$, <polar coordinates> give
$$
\begin{aligned}
f_*\Lambda(B)
&=\int_{\mathbb R^2}\mathbf1_B(x_1^2+x_2^2)\,dx_1\,dx_2\\
&=\int_0^{2\pi}\int_0^\infty\mathbf1_B(r^2)r\,dr\,d\theta\\
&=\pi\int_Bdt.
\end{aligned}
$$
The pushforward is therefore locally finite and atomless. By the <mapping theorem for Poisson point processes>, $f(\Pi)$ is a homogeneous <Poisson process> on $[0,\infty)$ with \b[constant intensity $\pi$]. This is the two-dimensional <radial volume transform of a homogeneous Poisson point process>.
Put $S_k=R_k^2$. The variable $S_k$ is the $k$th arrival time of that rate-$\pi$ Poisson process, hence has the <gamma distribution> with shape $k$ and rate $\pi$:
$$
g_k(s)=\frac{\pi^k}{(k-1)!}s^{k-1}e^{-\pi s},
\qquad s>0.
$$
Since $s=r^2$ and $ds/dr=2r$, the <change-of-variables formula for a probability density> gives the <Kth-nearest-neighbour distance in a homogeneous Poisson point process>
$$
\boxed{h_k(r)=2r\,g_k(r^2)
=\frac{2\pi r(\pi r^2)^{k-1}e^{-\pi r^2}}{(k-1)!},
\qquad r>0.}
$$
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