If $H(R)$ is spherically symmetric, we can write the Laplacian of $H$ in spherical coordinates:
$\nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg)$
(see wikipedia).
Then, $\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = R^2\frac{d^2H}{dR^2} + 2R \frac{dH}{dR}$
On the other way, $\frac{d}{dR}(RH(R)) = R\frac{dH}{dR} + H$
and thus
$\frac{d^2}{dR^2}(RH(R)) = R\frac{d^2H}{dR^2} + 2\frac{dH}{dR}$
$\Rightarrow \frac{1}{R^2}\frac{d}{dR}\bigg(R\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$
$\Rightarrow \nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$.
Now, apply it to $G_{\omega}$.