how to change metric variables The metric on unit sphere is given by:
$$g_{ij}=\begin{pmatrix}1 & 0 \\0 & \sin^2{\theta }\end{pmatrix}.$$
The laplacian beltrami operator in $\theta ,\phi$ is 
$$\Delta f=\left(\frac{1}{\sqrt{\vert g\vert}} \partial_{i} \ g^{ij}\sqrt{\vert g\vert}\partial_{j}f\right)$$
$$\Delta f=\left( \frac{\partial }{\partial\theta  }\sin(\theta)\frac{\partial}{\partial(\theta)}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \phi^2 }\right).$$
How to get these in $x,y$?
 A: There are probably more sophisticated (and perhaps "better") ways to do this, but I will stick to very elementary concepts:
The metric expresses how to get arc length.  If we embed the sphere into 3-space, the metric $(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2$ matches the given metric on the unit sphere.  But if we change $x$ and $y$, and stay on the sphere, the change in $z$ is determined:
$$
z = \sqrt{1-x^2-y^2} \\
dz = \frac{-x}{\sqrt{1-x^2-y^2}}dx + \frac{-y}{\sqrt{1-x^2-y^2}}dx
$$
where the two terms come from varying $x$ or $y$ by tiny amounts.  Note, by the way, that on the unit sphere equipped with $x,y$ by coordinates which are the projection in the $x,y$ plane of each point on the square, when you are at the equator, a tiny change in $x$ or $y$ forces a huge change in $z$.
We can read the metric from the expression for $(ds)^2$:
$$
(ds)^2 = (dx)^2 + (dy)^2 + \left( \frac{-x\,dx}{\sqrt{1-x^2-y^2}} + \frac{-y\,dy}{\sqrt{1-x^2-y^2}} \right)^2 \\
(ds)^2 = \left(1+ \frac{x^2}{1-x^2-y^2}(dx)^2\right)+ \left(1+ \frac{y^2}{1-x^2-y^2}(dy)^2 \right)
-\frac{2xy}{1-x^2-y^2}(dx)(dy)=
\frac{1}{1-x^2-y^2} \left( (1-y^2) (dx)^2 + (1-x^2) (dy)^2 - 2xy(dx)(dy)  \right)\\
g = \frac{1}{1-x^2-y^2} \pmatrix{1-y^2&-xy\\-xy&1-x^2}
$$
Note three points:  


*

*The sign of the $-xy$ terms.

*The fact that $||g||=1$ everywhere.

*The (removable) ring of coordinate singularity at $x^2+y^2 = 1$.


In fact, any metric imposed on the unit square will have a coordinate singularity somewhere.
