Elias M. Stein said that by an application of Green's theorem the following equality holds
$\int_{S^{n-1}}(\Delta _{S}f) gd\sigma=\int_{S^{n-1}}f( \Delta _{S}g)d\sigma$
where $\Delta _{S}$ is a spherical laplacian.
I don't know how to use the Green's theorem because both sides have the same domain.
The Laplacian is a second-order operator, so you'll apply Green's theorem twice. Since the sphere doesn't have boundary and the Laplacian is $\operatorname{div}\nabla,$
$$\int (\Delta f)g\ d\sigma = -\int \langle \nabla f,\nabla g\rangle\ d\sigma = \int f(\Delta g)\ d\sigma.$$
(Green's theorem is basically integrating by parts.)
Wikipedia calls this "Green's first identity." What I'm quoting is the $n$-dimensional analogue. Formally, the intuition is that you integrate by parts in each coordinate and since the boundary is empty, the boundary terms vanish: $$ \int f\Delta g = \int f\sum \partial_i^2g = \sum \int f\partial_i^2 g = -\sum\int \partial_if\partial_ig=-\int \langle\nabla f,\nabla g\rangle. $$
If you're comfortable with Stokes' theorem, this is a consequence of the product rule for the divergence: $$ 0 = \int_{\partial S} f\nabla g\cdot nd\sigma =\int_S \nabla \cdot (f\nabla g)d\sigma=\int_S\langle \nabla f,\nabla g\rangle +f\Delta g\ d\sigma $$