Current flow in a simple flyback topology
Now what I don't quite understand fully is what happens when the fet
  is turned off? I think the magnetic field will start to collapse and
  that this will result in current flowing in the other direction as the
  field collapses


No, this isn't what happens. The current doesn't change direction, the current stays the same and would gradually (if allowed) fall to zero dissipating the stored energy in other components.

So, in the beginning there was nothing then, suddenly, the FET turned on and current began to flow through the inductor starting at zero amps and rising to some positive value after a few microseconds. The ramp in current is Vs/Lp, that is supply voltage divided by primary inductance. Anything on the secondary doesn't count because of dot-notation and the diode in the secondary.

An amount of energy is stored in the magnetic field and when the FET open circuits that energy can only be passed through to the secondary and luckily (not really) the diode is going to conduct and pass that energy through to the load and storage capacitor.

Under no circumstances does the body diode of the FET conduct because the  flyback voltage at the drain is positive. It has to be positive (and remains positive) while ever the fet is open circuit. 

Just think about it - the long-term average voltage across an inductor has to be zero - if it isn't then the average current will be a rediculously large amount.