Modifications to 555 LED Fade Circuit
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 &amp; 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).