The line element is:

ds^2 = c^2(1-2m/r)dt^2 - (1-2m/r)^(-1)dr^2 - r^2 d theta^2 - r^2sin^2(theta)d phi^2

where m = MG/c^2

You know that in a flat space-time the trajectory of a photon is such that ds^2 = 0 on the trajectory (such a trajectory is called a null geodesic). This is also true in general relativity. It follows form the usual reasoning that you can always define a local "free falling" coordinate system in which the metric looks locally like that of special relativity. In that coordinate system ds^2 will be zero for two nearby points in the photon's trajectory, but because ds^2 is an invariant, it must be zero in the original coordinates too.

For a photon moving in the radial direction we then have:

0 = c^2(1-2m/r)dt^2 - (1-2m/r)^(-1)dr^2 --->

dr/dt = +/- c (1-2m/r)

For a photon moving from r1 to r2 we must choose the plus sign. We can write this as:

dt = 1/c dr/[1-2m/r] = 1/c rdr/[r-2m]

Integrating from r1 to r2 gives the coordinate time needed for the photon to go from r1 to r2

t_{1,2} = 1/c Integral from r1 to r2 of dr r/[r-2m] =

1/c Integral from r1 to r2 of dr (r-2m+2m)/[r-2m] =

1/c Integral from r1 to r2 of dr 1 + 2m/[r-2m] =

[r2 - r1]/c + 2m/c Log[(r2-2m)/(r1-2m)]

The total coordinate time needed for the photon to go from r1 to r2 and back to r1 is, of course:

Delta t = 2 t_{1,2}

If you consider the departure and arrival of the photon at position r1 as two events, then these two events are separated by a distance of zero and a coordinate time of Delta t. The space time interval between these two events is therefore:

Delta s^2 = c^2(1-2m/r)Delta t ^2

By the usual reasoning you identify Delta S ^2 as c^2 (Delta tau)^2 ,where Delta tau is the proper time that the clock measures. You then find that:

Delta tau = sqrt[1-2m/r1]Delta t