I think that the solution to your problem is to define the bins arrays for your histogram (for instance a linspaced array between 0 and 2pi for theta and between 0 and 1 for r). This can be done with the bins or range arguments of function numpy.histogram
I you do so, make sure that the theta values are all between 0 and 2pi by plotting theta % (2 * pi) instead of theta.
Finally, you may choose to plot the middle of the bin edges instead of the left side of the bins as done in your example (use 0.5 * (r_edges[1:] + r_edges[:-1]) instead of r_edges[:-1])
below is a suggestion of code
import matplotlib.pyplot as plt
import numpy as np
#create the data
r1 = .2 + .2 * np.random.randn(200)
theta1 = 0. + np.pi / 7. * np.random.randn(len(r1))
r2 = .8 + .2 * np.random.randn(300)
theta2 = .75 * np.pi + np.pi / 7. * np.random.randn(len(r2))
r = np.concatenate((r1, r2))
theta = np.concatenate((theta1, theta2))
fig = plt.figure()
ax = plt.subplot(111, polar=True)
#define the bin spaces
r_bins = np.linspace(0., 1., 12)
N_theta = 36
d_theta = 2. * np.pi / (N_theta + 1.)
theta_bins = np.linspace(-d_theta / 2., 2. * np.pi + d_theta / 2., N_theta)
H, theta_edges, r_edges = np.histogram2d(theta % (2. * np.pi), r, bins = (theta_bins, r_bins))
#plot data in the middle of the bins
r_mid = .5 * (r_edges[:-1] + r_edges[1:])
theta_mid = .5 * (theta_edges[:-1] + theta_edges[1:])
cax = ax.contourf(theta_mid, r_mid, H.T, 10, cmap=plt.cm.Spectral)
ax.scatter(theta, r, color='k', marker='+')
ax.set_rmax(1)
plt.show()
which should result as
