Making a voltage drop while providing high current - how does this work?
A voltage divider will not work.

If you dis want to do this with a voltage divider, then you would need to make sure that there was much more current through the divider than could ever be drawn by the chip.  Otherwise the chip will short out the lower half of the voltage divider, and the voltage will drop. But that would mean having e.g. 20A flowing through the divider, and even if USB could do that (it can't) it would make things very hot.

Use a low-dropout linear regulator, or, if accuracy is not important, a diode.

A linear regulator is designed to take a higher voltage and output a lower one.  It will adjust it's resistance to make that work.  You can buy one with a fixed output voltage of 4V, or you could buy an adjustable one and set the output with a couple of resistors.  A basic regulator will cost a few pence, if you want 2A you will need to spend more like £1.

A diode is simpler, and uses a little less space.  But it is a little trickier to get right, it needs a minimum current at all times to stop the voltage drifting up. So a linear regulator is probably easier to get right first time.

You will also want a capacitor next to the chip, to help smooth out those spikes.  The datasheet will have some suggestions - follow them. If the spikes are rare and short, a suitable capacitor or capacitors will also mean you cna get by with a smaller or cheaper linear regulator.