The derivation is similar to the one found at Hankel transform. Let $r=||\mathbf{r}||$, $k=||\mathbf{k}||$ and $F(\mathbf{k}):=\iint f(r) e^{i\mathbf{k}\cdot\mathbf{r}}\,d\mathbf{r}$. With no loss of generality, we can pick a hyperspherical coordinate system $(r, \theta,\ldots)$ such that the $\mathbf{k}$ vector lies on the $\theta = 0$ axis. The Fourier transform is now written in these hyperspherical coordinates as $F(\mathbf{k})=\int_{r=0}^\infty \int_{\theta=0}^{\pi}f(r)e^{ikr\cos \theta }\,r^{n-1}S_{n-2}(\sin \theta)\,dr\,d\theta$ Using $S_{n-2}(\sin \theta)=S_{n-2}(1)\sin^{n-2} \theta$ and changing integration order, we get $F(\mathbf{k})=\int_{r=0}^\infty f(r)r^{n-1}(S_{n-2}(1)\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,\sin^{n-2} \theta\,d\theta)\,dr$.
So $H(kr)=S_{n-2}(1)\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,\sin^{n-2} \theta\,d\theta$. Note that it would be better to write $H_n(kr)$ instead of $H(kr)$. For $n=2$, we have $S_0(1)=2$ and $\int_{\theta=0}^{\pi}e^{ikr\cos \theta }\,d\theta=\pi J_0(kr)$, so we get back the well known Fourier-Bessel transform.