Currents when converting voltages down
It depends on how the conversion is performed, and broadly there are two options:


linear converters have approximately the same input and output currents,
switching converters (and transformers) act like gearboxes, providing a different voltage/current ratio (which is like a torque/speed ratio) but still being limited by conservation of energy.


A linear converter drops the voltage across a semiconductor (transistor of some sort), therefore that transistor will dissipate a lot of heat.  If you have a 5V linear regulator supplied from your 12V 6A supply, then it too is capable of only 6A output and the regulator will dissipate $$P = (V_{in} - V_{out}) * I_{out} = (12-5)*6 = 42W$$
and that process has an efficiency of $$ n = P_{out} / P_{in} = (5*6)/(12*6) = 41.7\% $$

For small voltage drops like 1-3V this is a reasonable approach.  For larger drops though, you can see it becomes very inefficient because the efficiency is the voltage ratio.  Say you converted 120V to 12V linearly, that would be 10% efficient; the power supply would dissipate 9x as much as the load!

With a switching converter, conservation of energy means that $$ (P_{out} = V_{out} * I_{out}) \le (P_{in} = V_{in} * I_{in} )$$

Your 12V 6A supply has extremely poor efficiency: $$ n = (6*12)/(1.5*120) = 40\% $$
which leads me to believe that the 1.5A rating is for inrush current, not steady-state operation, or that the supply has very poor power-factor.  It's probably a transformer+rectifier+capacitor arrangement instead of a switching supply.

Anyway, say you had an 80%-efficient switching DC-DC converter, which is not unreasonable.  If you're powering it from a 12V 6A supply and want 5V from it, then: 
$$ P_{out} = V_{out}*I_{out} = n * P_{in} $$
$$ P_{out} \le 0.8*12*6 $$
$$ I_{out} = P_{out} / V_{out} \le 11.52A $$

So yes, you can get more current out by converting down to a lower voltage if you use the right kind of conversion.

Edit: if you're mechanically inclined then think of a voltage = torque and current = speed equivalence.
$$ power = voltage * current = torque * speed $$ 
An efficient voltage conversion process is just like a gearbox in that it allows you to have higher torque at lower speed or vice versa.

Edit 2: a linear conversion is like applying a brake.  It reduces the available output torque (voltage) by dissipating a bunch of power as heat, and doesn't change the speed (current) of operation.