I think it is because of low resistance of fan motor. I do not know what is the exact value but fan has 40A fuse so allowing for some reserve margin say if the fan draws 30A at 12V it would be 0.4 ohms. This resistance will vary depending on rpm as the inductive resistance will be added as well but for sake at argument lets assume resistance will be 0.5 ohm on stationary motor at the moment when power is first applied.
I still need to verify the voltages coming from the panel at good sunshine but lets say panel produces 8V load voltage and gives 3A of current for a total of 24W at best case scenario. If you put 0.5 ohm conductor at the output of panel it will try to pull 16A from it but as the panel can supply only 3A max the voltage will drop dramatically and overall output will be reduced.
I have modified the previous picture to show what I mean.
You would go from point A producing 24W to point B producing only 4.5W with 3A @ 1.5V. DC-DC converter with 90% efficiency allows to take that 8V 3A input to convert to 3.29V and 6.58A (21.6W which is a lot better than 4.5W when hooked up directly). Or if the sunshine can only produce 10W of power at 8V and 1.25A it would go down to 1.25A and 0.62V at 0.5 ohm resistance for 0.77W, DC-DC converter can convert it to 2.12V and 4.24A for 9W.
This is very simplified version with fan motor stationary with the least electrical resistance. As it starts to spin up inductive resistance will get higher so DC-DC converter output voltage will go up and current down maintaining the power.
