What exactly is a diode and how do I use it to step down voltage?
The suggested diode method cuts off another direction current in the alternating current circuit. That reduces the effective power 50% because half of the time the current is is blocked as effectively as the wire was disconnected. As commented, that's not 50% reduction of the effective voltage, it's less because the power is proportional to the square of the voltage in resistors. 

You probably do not see anything more than some brightness loss because the frequency probably is 60Hz where you live and every bulb will get current half of every 1/60th second. The filaments in the bulbs do not get cold that fast.

To keep the brightness intact you should remove half of the bulbs of one series if you add a diode. It's difficult to pretend how the lifetime of the bulbs suffers due the changed shape of the current pulses. 

The diode must stand the peak AC voltage and the current. 

The diode does not get hot because it works like a switch. That's true, if the diode is selected for the load current and can be cooled freely by the air.

Serial resistor doesn't block the current totally, both halves of the AC get through, but part of the total power isn't dissipated in bulbs, it's dissipated in the resistor. 

The resistor generates heat. If the resistor is selected properly using Ohm's law it dissipates electric energy to heat exactly as much as the bulbs which are replaced by that resistor. If  the resistor has about the same dimensions as one bulb (=same ability to get cooled by the air), but dissipates worth of ten bulbs, it surely gets hot because its temperature must rise 900% higher over the ambient temperature than one bulb. 

This text is valid only for incandescent bulbs which have a filament. Led lights work totally differently.