Power Supply for 555 Timer LED Fade In/Out

I suspect that they use a 9 volt battery since it is a handy, readily-available power source.

The electrical characteristics of LEDs (and diodes in general) are a bit odd - they don't follow Ohm's Law. The voltage across an ordinary ("bare") LED depends primarily on its colour, and will vary only slightly with current (for a red LED, the voltage will be about 1.8 volts). Therefore it is necessary to include something in an LED circuit to limit the current to a safe value - the 1K resistor in the circuit you linked to serves as the currrent limiter.

An LED advertised as "12 volts" or "5 volts" will include a series resistor or other current-limiting device, and can be connected directly to the advertised voltage with no other components.

A "bare" red LED will have a voltage drop of about 1.8 volts, yellow is about 2 volts, and green is about 2.2 volts. Blue and white are about 3.3 volts (white LEDs are really blue LEDs with a yellow phosphor).

A simple analysis of the original circuit and the substitutions you made goes something like this.

The 555 circuit shown is a squarewave (1:1) oscillator. The frequency being determined by R1 and C1. The voltage at the junction between these two components changes between 2/3 rds and 1/3rd of the supply (set by input pins 6 and 2). For a 9V supply this will be 6V & 3V. Changing the supply to 12V gives 8V and 4V. The voltage varies in an approximately triangular waveform.

This voltage is connected to the the base of an NPN transistor, Q1 (in this case a BC478.) Just about any small signal NPN would do so a change to a 2N2222 type would have very little effect on the operation. The transistor is connected as an emitter follower so the voltage at the emitter would be 0.6V lower than its base voltage. For a 9V supply this voltage would be between 5.4 and 2.4V. For a 12V supply this would be between 7.4V and 3.4V.

The 556 is just a double 555 in one package so using one half of the chip should still work (provided you have the correct connections).

Your last substitution is where it all goes wrong - using a 12V LED (with built in resistors.)

A normal (visible) LED needs between 2 - 4 Volts to turn on. This voltage depends upon colour (i.e. the energy gap). R3 is chosen to limit the maximum current through the LED (and transistor). The 9V circuit matches the 5.4V - 2.4V range very nicely for something like a RED LED (about 2V). For the 12V supply you would be better off using a COOL WHITE or BLUE LED (about 3.8V). The R3 value of 470R is fine. For a 12V circuit and WHITE LED (3.8V turn on) this gives a maximum current of (7.4 - 3.8)/470 A or about 7.7mA. Compared with (5.4 - 2)/470 A or about 7.2mA (RED LED, 2V turn on).