= Solution
For the <spherical coordinates> in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin\theta$. Their <metric tensor> is therefore $\operatorname{diag}(1,r^2,r^2\sin^2\theta)$ and the <Jacobian determinant> is $r^2\sin\theta$. Applying the divergence formula to the <gradient> gives
$$
\boxed{\Delta u=\frac1{r^2}\partial_r(r^2u_r)+\frac1{r^2\sin\theta}\partial_\theta(\sin\theta\,u_\theta)+\frac1{r^2\sin^2\theta}u_{\phi\phi},\quad dx=r^2\sin\theta\,dr\,d\theta\,d\phi.}
$$
For example, the formula follows directly from $\Delta u=|\det h|^{-1/2}\partial_i(|\det h|^{1/2}h^{ij}\partial_ju)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the <Laplacian> is a smooth geometric operator there as well, described by other charts. The coordinate <volume form> integrates over $1/2<r<2$, $0<\theta<\pi$, $0\le\phi<2\pi$, with the usual periodic identification.
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