Why is light described by a null geodesic?

Even in curved spacetime, you can perform a coordinate transformation at any location ("move to a freely falling frame") such that your metric is locally flat , and takes the form \begin{equation} ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2\end{equation}

If you consider a null trajectory where $ds^2 = 0$, then the above equation takes the form

\begin{equation} cdt = \sqrt{dx^2 + dy^2 + dz^2}. \end{equation}

This is the statement that "the speed of light times the differential time interval, as measured by an observer in a freely falling frame at the location in consideration, is equal to the differential physical distance traveled along the trajectory, measured by that same observer." From Einstein's equivalence principle, this is precisely the way that light must behave.

Suppose there is a flash event that we can represent as a light cone as the flash expands.

There are three shutters with detectors around this flash event. The shutters open and close only once and this is almost instantaneous. One shutter opens in spacetime outside the light cone. One Shutter opens in spacetime inside the lightcone and the third opens exactly on the edge of the light cone.

The first shutter misses the flash because it opens before the flash reaches the shutter. According to the spacetime diagram the separation between the flash and the shutter is spacelike because there is a distance of space between the shutter opening and the flash reaching the shutter. The second shutter misses the flash because the flash has already passed over it in the past. Thus the separation between the shutter and the flash is timelike. The shutter that catches the flash has no separation. There is no difference in space or time between the shutter opening and the flash entering the shutter. You could say the difference is Null. Thus you can think of it as a null geodesic.