This is going to be unsightly...

The following Mathematica code:

<pre>
Needs["VectorAnalysis`"]
Simplify@ CoordinatesFromCartesian[
 CoordinatesToCartesian[{r, theta, phi}, Spherical] 
                     + CoordinatesToCartesian[{r0, theta0, phi0}, Spherical],
 Spherical
 ] 
</pre>

gives the following output (doctored so that it looks nicer):

$$ r' = \sqrt{r^2+2 r_0 r \left(\sin (\theta ) \sin
   \left(\theta _0\right) \cos \left(\phi -\phi
   _0\right)+\cos (\theta ) \cos \left(\theta
   _0\right)\right)+r_0^2} $$

$$ \theta' = \cos ^{-1}\left(\frac{r \cos (\theta )+r_0 \cos
   \left(\theta _0\right)}{\sqrt{r^2+2 r_0 r
   \left(\sin (\theta ) \sin \left(\theta
   _0\right) \cos \left(\phi -\phi
   _0\right)+\cos (\theta ) \cos \left(\theta
   _0\right)\right)+r_0^2}}\right) $$

$$ \phi' = \tan ^{-1}\left(r \sin (\theta ) \cos (\phi
   )+r_0 \sin \left(\theta _0\right) \cos
   \left(\phi _0\right),r \sin (\theta ) \sin
   (\phi )+r_0 \sin \left(\theta _0\right) \sin
   \left(\phi _0\right)\right) $$

In this last line, there is a two-argument variant of arctan, which is explained [here](http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent), for example.