Ok, I think I understood your question. You started from $dl^2=dx^2+dy^2+dz^w+\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$ and tried to pass to spherical coordinates, as it is done on the wiki page.
Set $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, then the Jacobian matrix is $J=\begin{pmatrix}\sin\theta\cos\phi&r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\ \sin\theta\sin\phi&r\cos\theta\sin\phi&r\sin\theta\cos\phi\\\cos\theta&-r\sin\theta&0\end{pmatrix}$ so the metric $dx^2+dy^2+dz^2$ in these new coordinates is given by $g=J^tJ=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2\theta\end{pmatrix}$ i.e. $dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2$.
So $dl^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2+\frac{r^2dr^2}{\kappa^{-1}R^2-r^2}$ from which $dl^2=dr^2\left(1+\frac{1}{\frac{R^2}{\kappa r^2}-1}\right)+r^2d\theta^2+r^2\sin^2\theta d\phi^2$ and $1+\frac{1}{\frac{R^2}{\kappa r^2}-1}=\frac{\frac{R^2}{\kappa r^2}}{\frac{R^2}{\kappa r^2}-1}=\frac{1}{1-\frac{\kappa r^2}{R^2}}$ so, plugging this into the previous, you obtain $dl^2=\frac{dr^2}{1-\frac{\kappa r^2}{R^2}}+r^2d\theta^2+r^2\sin^2\theta d\phi^2\;.$ Does this answer to your doubt?