tapped inductor buck converter
Tapped inductor buck solves the problem of low duty-cycle at high input voltages. As you probably know, the duty-cycle of a regular buck is given by the output voltage divided by the input voltage (approximately, the exact formula is a little more complicated). As you can imagine, for a buck that works up to 265VAC (375VDC), with an output voltage of say, 6VDC for 3 LEDs in series, the duty-cycle would become 6/375=0.016. At 100kHz, the switch would be on for only 160ns. That is not practical.
The calculation of the duty-cycle for the buck is governed by Faraday's law. That requires that the volt*seconds applied to the inductor during the on time be equal to the volt*seconds during the off time. In a regular buck the voltage across the inductor is clamped by the catch diode to the output voltage (actually output voltage plus the catch diode drop; this is one of the corrections to the duty-cycle formula: DC≈(Vout+Vdiode)/(Vin+Vdiode) ).
With the tapped buck the voltage is clamped to the output voltage only across the section of the inductor that is connected to the catch diode. The voltage across the other section of the inductor will be simply the output voltage multiplied by the turns ratio of the two sections. The inductor behaves like a transformer now.
All this means the voltage across the inductor during the transistor off time is no longer limited to the output voltage plus one diode drop, but it is much larger, it is the output voltage plus diode drop times 1 plus the turns ratio
Voff=(Vout+Vdiode)*(1+N)
With this, the on time can be much longer and it can become practical. That is why you have the tapped inductor buck.
Regarding your second question, no the voltage is not calculated that way. As I said above, the inductor behaves like a transformer and it behaves like a flyback transformer. So the voltage across the feedback winding will be the output voltage plus a diode drop times the turns ratio between the output section and the feedback section. The current in the feedback winding will then be that voltage divided by the value of the resistors.
The main thing to remember is that it works much like a flyback and the output section of the winding is the one that clamps the voltage and the other windings just follow that with their respective turns ratios.
One thing I think is missing in that schematic is a clamp for the transistor. As in any transformer, there will be leakage between the sections of the inductor, causing leakage spikes. These can destroy the switcher. So if you are going to build it, add a snubber of some sort to the drain of the switcher.