= Solution
For $1\leq k\leq n$, the <nearest neighbour distance> $\rho_{(k)}(x)$ is the $k$th <order statistic> of $D_i=|X_i-x|$. Each $D_i$ has continuous <distribution function> $p_x(r)$ on $[0,\infty)$. The <probability integral transform> makes $U_i=p_x(D_i)$ <independent and identically distributed random variables> with the <uniform distribution> on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a <uniform order statistic>. Strict monotonicity is not needed at this stage.
For $0<s<1$, counting observations below $s$ gives
$$
\mathbb P(P\leq s)=\sum_{j=k}^n\binom nj s^j(1-s)^{n-j}.
$$
Differentiating and cancelling consecutive terms gives the <Beta distribution>:
$$
\boxed{f_P(s)=\frac{n!}{(k-1)!(n-k)!}s^{k-1}(1-s)^{n-k},\qquad 0<s<1.}
$$
The <Gamma function> identity $\Gamma(j)=(j-1)!$ identifies this with the stated normalization. Ratios of <Beta function> integrals yield the <moments>
$$
\boxed{\mathbb EP=\frac{k}{n+1},\qquad \mathbb EP^2=\frac{k(k+1)}{(n+1)(n+2)}.}
$$
This is the <probability content of a nearest neighbour ball> identity, specialized to intervals on the real line.
Now write $f_0=f(x)>0$. The <Lipschitz bound> gives
$$
|p_x(r)-2rf_0|
=\left|\int_{-r}^r\{f(x+u)-f_0\}\,du\right|
\leq L\int_{-r}^r|u|\,du=Lr^2.
$$
Thus
$$
\boxed{|p_x(r)-2rf(x)|\leq Lr^2.}
$$
Strict positivity of the <probability density function> makes $p_x$ continuous and strictly increasing from zero to one, so its <inverse function> is defined for every $s\in(0,1)$.
For the requested conditional calculation, suppose $f_0^2\geq L$. At $r_0=s/f_0$ the preceding bound gives
$$
p_x(r_0)\geq2s-\frac{Ls^2}{f_0^2}\geq2s-s^2\geq s.
$$
Therefore $r=p_x^{-1}(s)\leq s/f_0$, and substituting in the approximation proves the <inverse probability content bound for a Lipschitz density>:
$$
\boxed{|2f(x)p_x^{-1}(s)-s|\leq\frac{Ls^2}{f(x)^2}.}
$$
The same argument works, without $f_0^2\geq L$, on the meaningful local range $0<s<1$ with $s\leq f_0^2/L$.
Since the <nearest neighbour density estimator> obeys $f(x)/\widehat f_{(k)}(x)=2(n+1)f_0p_x^{-1}(P)/k$, its conditional bound follows from the <Beta distribution> <moments>:
$$
\begin{aligned}
\left|\mathbb E\left[\frac{f(x)}{\widehat f_{(k)}(x)}\right]-1\right|
&=\frac{n+1}{k}\left|\mathbb E\{2f_0p_x^{-1}(P)-P\}\right|\\
&\leq\frac{n+1}{k}\frac L{f_0^2}\mathbb EP^2
=\frac L{f_0^2}\frac{k+1}{n+2}.
\end{aligned}
$$
Hence, under the printed location condition,
$$
\boxed{\left|\mathbb E\left[\frac{f(x)}{\widehat f_{(k)}(x)}\right]-1\right|\leq\frac{k+1}{n+2}.}
$$
The conditional <inverse function> bound also bounds $p_x^{-1}(P)$, so the <expected value> in this calculation is finite under those formal hypotheses.
There is, however, a genuine flaw in the original PDF: \b[the printed condition $f(x)\geq\sqrt L$ has no admissible points when $f$ is strictly positive on all of $\mathbb R$]. A <probability density function> on all of $\mathbb R$ cannot have <Lipschitz constant> zero, so $L>0$. Its <Lipschitz bound> forces
$$
f(x+u)\geq\max\{f_0-L|u|,0\}.
$$
Integrating this triangular lower envelope gives the <Lipschitz density height bound>
$$
1=\int_{\mathbb R}f(y)\,dy>\int_{-f_0/L}^{f_0/L}(f_0-L|u|)\,du=\frac{f_0^2}{L}.
$$
The inequality is strict because the <probability density function> has positive mass outside the finite interval. Thus $f(x)<\sqrt L$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\leq\sqrt L$, with equality forcing the entire <probability density function> to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final <expected value> bound, since $P$ ranges over all of $(0,1)$.
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