[This answer doesn't add much that is new, but it became too painful trying to post is as a comment.]
One doesn't need to talk about differential forms, or integral identities, in order to prove the validity of vector calculus formulas in spherical coordinates (although both those points of view are very nice!). If $\vec{F}(x,y,z)$ is a vector field expressed in Cartesian coordinates, say, and $\vec{F}(r,\theta,\varphi)$ is the same vector field but now expressed in spherical coordinates, then the formula for $\nabla$ in spherical coordinates is determined by the requirement that $\nabla\cdot \vec{F}(x,y,z)$ and $\nabla\cdot \vec{F}(r,\theta,\varphi)$ correspond to one another as vector fields when you change from Cartesian to shperical coordinates.
Similarly, if $\vec{A}(x,y,z)$ and $\vec{B}(x,y,z)$ are a pair of vector fields expressed in Cartesian coordinates, with $\vec{A}(r,\theta,\varphi)$ and $\vec{B}(r,\theta,\varphi)$ being the same vector fields but now expressed in spherical coordiantes, then $\vec{A}\times \vec{B}$ (expressed in Cartesian coordinates) will correspond to $\vec{A}\times
\vec{B}$ (expressed in spherical coordinates). So the left-hand side of the identity to be checked, when computed in Cartesian coordinates, will match (under the change from spherical to Cartesian coordinates) with the same
expression when computed in spherical coordiantes. Similarly for the right-hand side of the identity. Thus the identity is equally valid in Cartesian or spherical coordinates. (Again, the key point is that the formulas for grad, div, and curl under a coordinate change are defined so as to make the above argument valid.)
Here's what's happening in $\mathbb{R}^3$ with rectangular coordinates. You can tweak as needed.
Let $g(x,y,z)$ be a smooth scalar function and $\mathbf{F}(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$ be a smooth vector field. Then
\begin{align}
\nabla \cdot (g\,\mathbf{F})&=\nabla\cdot((gF_1,gF_2,gF_3))\\
&=(gF_1)_x+(gF_2)_y+(gF_3)_z\\
&=g_xF_1+g(F_1)_x+g_yF_2+g(F_2)_y+g_zF_3+g(F_3)_z,
\end{align}
while
\begin{align}
\nabla g\cdot \mathbf{F}&=g_x F_1+g_y F_2+g_z F_3,\\
g\,(\nabla\cdot \mathbf{F})&=g\,((F_1)_x+(F_2)_y+(F_3)_z)=g(F_1)_x+g(F_2)_y+g(F_3)_z.
\end{align}
Adding these last two yields the first.
Best Answer
HINT: Use identity 7 with $\mathbf{F}=\nabla g\times\nabla h$.