The rule you have for the total $\Delta V$ of a low-thrust spiral is an upper limit arrived at as you let the thrust go to zero. However that takes an infinite amount of time. The total $\Delta V$ of a spiral with non-zero thrust is less, and the time is finite. But it is a good rule of thumb for quick calculations when trying to establish feasibility.
The derivation of the rule of thumb is quite simple. Look at an infinitesimally small Hohmann transfer. You will find that the $\Delta V$ total of the two infinitesimal burns at the initial orbit and at apoapsis of the transfer orbit is equal to the difference in the orbital velocities. Then if you add those up for a finite raise in orbit, you get the difference in $\Delta V$ of the initial and final orbit.
To find out the real total $\Delta V$ and to plot an actual trajectory that doesn't do an infinite number of orbits before it gets anywhere is best done using numerical integration.
Here is an example of a spiral from a circular orbit to escape ($C_3=0$):
This is normalized to the starting circular orbit, where the distances are in units of the initial orbit radius, and the acceleration is constant at $10^{-3}$ of the gravitational acceleration of the body at the initial orbit radius. The total $\Delta V$ to escape is 0.856 of the initial orbit velocity, as compared to 1.0 for the rule of thumb. The total time to escape is 136 initial orbit periods. It goes around the body about 40 times before escaping.
The first several orbits are close enough that you can't make them out at the resolution shown. This gets even worse for smaller accelerations. $10^{-3}$ is actually pretty high. I picked it so that you can see the spiral better. That time from a low Earth orbit is about 8.5 days. A typical spiral out might be more like months with accelerations of $10^{-4}$ of the initial gravitational acceleration, or less. Attempts at plotting that show a solid disk until near the end where you see the spiral escape.
Here is an example of a spiral from LEO (400 km) to GEO with the same normalization and a normalized constant acceleration of $10^{-4}$. It takes about two months over 945 orbits. In this case the total $\Delta V$ is very close to the rule of thumb. This is simplified, since the final flight path angle here is about half a degree. So there is some time and $\Delta V$ remaining to circularize the orbit.
You could approximate this plot by advancing one orbit at a time, using the orbit period times the acceleration as the $\Delta V$ and raising the orbit the corresponding amount, connecting each with a linearly increasing spiral.