Mapping Coordinates on a Plane Tangent to a Sphere into Cartesian Coordinates in 3D Space Before we begin, I must ask you to keep the vocabulary at high school level.
These variables define the point the plane needs to be tangent to – the center of the circle is at the origin.
r - defines the radius of the sphere
Ø1 – defines the rotation in the x-z axis – can be from 0 to 2π
Ø2 – defines the rotation in the y axis – can be from 0 to π  


I have found that I can convert that point on the sphere to Cartesian coordinates using equations.
x0=r•cosØ1•sinØ2
z0=r•sinØ1•sinØ2
y0=r•cosØ2
These variables define the location of the point to be converted on the 2d plane that is tangent to the sphere.
d1 – is the x coordinate
d2 – is the y coordinate  
Positive y for the 2d plane is in the vertical direction as much as possible (because the plane is tangent to the sphere), positive x is in the counterclockwise direction looking down on the sphere from positive y in 3d space (or to the right if the plane is on your side of the sphere).

The formulas I have made are:
x1=x0+d2•cosØ1•cosØ2-d1•sinØ1
z1=z0+d2•sinØ1•cosØ2+d1•cosØ1
y1=y0+d2•sinØ2
At this point I hope you understand what I am trying to do and can point me to where I’ve gone wrong.
 A: The first thing I did is I changed the axis to a more conventional format.

All I really did is switch the y and z axis.
To that effect, this is the revised polar-to-cartesian coordinate conversion equations.
x0=r•cosØ1•sinØ2
y0=r•sinØ1•sinØ2
z0=r•cosØ2
This is the plane that is tangent to the sphere.

These are the equations that convert coordinates on the plane into 3d cartesian coordinates.

And the written version (in case you can't load images for some reason)
x1=x0-d2•cosØ1•cosØ2-d1•sinØ1
y1=y0-d2•sinØ1•cosØ2+d1•cosØ1
z1=z0+d2•sinØ2 
This is a close-up on the diagrams.

A: So you are asking to find the equation of the plane (or points $(x_1,y_1,z_1)$ on the plane) that is tangent to the sphere at the point $(x_0,y_0,z_0)$. 
I will following the regular right hand rule, so the $z$ axis in your first picture should point outward. The $\theta_2$ starts from the positive $y$ direction.
So your $(x_0,y_0,z_0)$ in spherical coordinates is correct. Then you find the directive with respect to $theta_1,theta_2$ to find the tangent direction. 
The new points are 
$$ x_1=x_0+d_2\cos{\theta_2}\cos{\theta_1}-d_1\sin{\theta_2}\sin{\theta_1}\\
 z_1=z_0+d_2\cos{\theta_2}\sin{\theta_1}+d_1\sin{\theta_2}\cos{\theta_1}\\
 y_1=y_0-d_1\sin{\theta_2}$$
