As the paper is quite long and sometimes technical, here's the gist of the calculation:
Assuming we've somehow dealt with the infinite extension of the plane and the resulting unnormalizability of distributions over "all possible configurations" (which the paper does), we can say that the "distribution" of relative configurations of $3$ points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is uniform with respect to the volume element in the four-dimensional space of Cartesian coordinate differences $(\Delta x_2,\Delta y_2,\Delta x_3,\Delta y_3)=(x_2-x_1,y_2-y_1,x_3-x_1,y_3-y_1)$. We can then transform to the "circumdisk representation" $(x_i,y_i)=(x_0,y_0)+R(\cos\theta_i,\sin\theta_i)$, where the circumcentre $(x_0,y_0)$ cancels in the coordinate differences, and calculate the Jacobi determinant:
$ \begin{eqnarray} \left| \frac{\partial(\Delta x_2,\Delta y_2,\Delta x_3,\Delta y_3)}{\partial(R,\theta_1,\theta_2,\theta_3)} \right| &=& \left| \begin{array}{cccc} \cos\theta_2-\cos\theta_1&R\sin\theta_1&-R\sin\theta_2&0\\ \sin\theta_2-\sin\theta_1&-R\cos\theta_1&R\cos\theta_2&0\\ \cos\theta_3-\cos\theta_1&R\sin\theta_1&0&-R\sin\theta_3\\ \sin\theta_3-\sin\theta_1&-R\cos\theta_1&0&R\cos\theta_3\\ \end{array} \right| \\ &=& R^3\left(\sin(\theta_1-\theta_2)+\sin(\theta_2-\theta_3)+\sin(\theta_3-\theta_1)\right)\\ &=&\pm4R^3\sin\alpha\sin\beta\sin\gamma\;, \end{eqnarray} $
where $\alpha,\beta,\gamma$ are the angles of the triangle formed by the three points. Thus the "distribution" factorizes into radial and angular parts. As Michael pointed out in a comment under the answer on MO, the angular part is proportional to the ratio of the area of the triangle to the area of the circumcircle, so the more of its circumcircle it occupies, the more likely a triangle is to appear.
Note that this is all about general triangles and doesn't refer to Delaunay triangles yet, so for instance if we consider all triangles with circumradii in a certain range, their angles are distributed according to $\sin\alpha\sin\beta\sin\gamma$. Since a triangle formed by any three points is a Delaunay triangle iff its circumcircle doesn't contain any other points, the distribution for Delaunay triangles is proportional to the above result times the exponential factor $\exp (-\pi\rho R^2)$ (with $\rho$ the intensity of the Poisson process) to account for the probability of the circumcircle being empty.
Incidentally, if we follow Michael's first prescription and consider the distribution for the triangle surrounding the origin, we have to multiply by the same area factor again, so the angular distribution in this case is proportional to $\sin^2\alpha\,\,\sin^2\beta\,\,\sin^2\gamma$.