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This is the Laplace operator of Spherical coordinates:
What is the Laplace operator of Schwarzschild-Spherical coordinates?
where the Differential displacement of Schwarzschild-Spherical coordinates is:
$$d{\bf{l}}=(1-\frac{r_s}{r})^{-1/2}dr\hat{\bf{r}}+rd\theta{\bf\hat{\theta}}+\sin{\theta}rd\phi{\bf\hat{\phi}}$$
It seems you want to calculate the laplacian in the spacial surface of $t$ constant. If the global metric is defined as:
$$ ds^2 = (1-\frac{r_s}{r}) \, dt^2 + (1-\frac{r_s}{r})^{-1} \, dr^2 + r^2 \, d\theta^2 + r^2 \, \sin^2(\theta) \, d\phi^2 $$
The metric on the surface $t$ constant is $$ ds^2 = (1-\frac{r_s}{r})^{-1} \, dr^2 + r^2 \, d\theta^2 + r^2 \, \sin^2(\theta) \, d\phi^2 $$
So your metric written in matrix form is: $$ [g_{ij}]=\begin{pmatrix} (1-\frac{r_s}{r})^{-1} && 0 && 0 \\ 0 && r^2 && 0 \\ 0 && 0 && r^2 \, \sin^2(\theta) \end{pmatrix} $$ The laplacian operator can be written in terms of this matrix elements as other people pointed in the comments. The general formula can be found in wikipedia as $$ \nabla^2 \phi = \sum_{ij} \frac{1}{\sqrt{|\det(g)|}} \partial_i \left(g^{ij} \sqrt{|\det(g)|} \partial_j \phi \right) $$ Where $g^{ij}$ is the inverse of the matrix of the metric $g_{ij}$. So $$ [g^{ij}]=\begin{pmatrix} (1-\frac{r_s}{r}) && 0 && 0 \\ 0 && \frac{1}{r^2} && 0 \\ 0 && 0 && \frac{1}{ r^2 \, \sin^2(\theta)} \end{pmatrix} $$ Now you can use the formula above for get the laplacian you want.
