Question: I'm trying to understand ways of isolating circuits and general concepts of isolation.

* I understand how high frequency signals can pass over a capacitor.
* I understand also how optocouplers work.
* I understand how transformers work.

Many places talk about using transformers to isolate,but as far as I know this is only for AC sources. They talk about using transformers for isolation but don't take into account how to isolate DC power sources.

How I can isolate a 48 V DC rail that feeds a DC-DC buck regulator with 12 V DC output? I can think of no way to transfer a large amount of DC energy without direct contact.

I understand induction, electromagnetic waves, capacitance, optocoupling, saw waves but I don't know why all this is related to galvanic isolation. Why galvanic? If for example I use 220 V AC isolation transformer, where is the galvanic current, or the galvanic "phenomena"?

Answer: 
In a word, no. The FFC cable would likely overheat and maybe melt or burn.

Both your connector and FFC cable are rated 500mA max.

Therefore if you have five separate 500mA constant current sources, you can't combine their return paths to a single wire, as it would mean 2.5A return current through single wire.

You have to have 5 return paths. Either separate return paths, or, if you combine the return paths on the LED board, several ground wires on the FFC is still needed, then a minimum of 5, preferably more, because if one or a few ground return wires make a bad connection, then all the return current would exceed the current carrying ability of the remaining wires.
The FFC contacts are rated for current by each supplier. Not all contacts have the same area.

e.g. 1mm pitch 1.0 A