Interpreting bode plot of type 3 compensation network for buck regulator
Calculated values for the poles and zeros look about right, except for the missing integrator pole at the origin. Maybe it would be best to break the loop up into pieces. Write the transfer function of the error amp, from \$V_{\text{out}}\$ to error amp output to start. You should get something like: 

\$\frac{\text{EA}_{\text{out}}}{V_{\text{out}}}\$ = \$\frac{\text{R2 } \text{EA}_{\text{gm}}}{\text{R1}+\text{R2}}\$ \$\frac{\text{Cs } \text{Rs } s+1}{s (\text{Cp}+\text{Cs}) \left(\frac{\text{Cp } \text{Cs } \text{Rs } s}{\text{Cp}+\text{Cs}}+1\right)}\$ 

Which for the values shown in the schematic (except, using R1=10kOhms and R2=1kOhm, since values for those were not shown) would look like: 



Then write the output filter response. It would be like this: 

\$\frac{V_{\text{out}}}{V_{\text{sw}}}\$ = \$\frac{\text{Co } \text{esr } \text{Ro } s+\text{Ro}}{s^2 (\text{Co } \text{esr } \text{Lo }+\text{Co } \text{Lo } \text{Ro })+s (\text{Co } \text{esr } \text{Ro }+\text{Lo})+\text{Ro}}\$ 

Which would look like this: 



Of course the total loop would also have the modulator gain which would be ~ \$V_{\text{in}}\$ / \$V_{\text{ramp}}\$. Also left out are the Nyquist poles.

You can see that the loop put together will be unstable unless another zero is added to the error amp response. It might also be necessary to damp the LC filter since with such a small amount of esr the Q is quite high. Under damped filters are a common problem with this type of regulator. 



About Infinite Gain Margin, and what it means when Phase doesn't cross -180. 

I don't think infinite gain margin is meaningful or realizable. A situation like that shows a limitation or inaccuracy of the model: Poles or loop delays that haven't been included, like Nyquist poles and higher frequency poles that exist in the error amp (both of which have been left out here). Or, it might show an error of interpretation, that the plot is not of Phase, but Phase margin, which wouldn't cross -180 but instead 0 degrees.