LM1084 , LT1084, *1084 ... linear regulators capable of up to 5A and with a low dropout voltage (1v, up to 1.4v depending on temperature, extreme case) and they can be paralleled with a very small resistor to balance them.
See
http://www.ti.com/lit/ds/symlink/lm1084.pdf or
http://cds.linear.com/docs/en/datasheet/108345fh.pdf (see page 15 for paralleling regulators)
LM1084 is cheap, about 2-3$ for one piece, the LT1084 is more expensive, as Linear parts often are (but if you're super cheap you can request samples from them - up to 2 parts per product, up to 3 products per request if I remember correctly - and you'll get them by mail in a couple of weeks)
If you want 6v out using a linear regulator like the one above, you'd need a minimum voltage always above 6v + dropout voltage , or about 7v. It's in your best interest to keep the minimum and maximum voltage these linear regulators see as close as possible to that 7v figure, because the difference between input and output voltage is dissipated as heat and no matter what heatsinks you use, the maximum you'd dissipate on the regulators would be about 15-20w (see note 4 on page 6 of Linear's datasheet for maximum dissipated power depending on regulator and the case)
So for example, if you'll use two regulators in parallel and each of them will output 4 amps (out of maximum 5 amps), that 20w figure means you have a budget of 20w / 4 amps = 5v , which means you should not exceed 6v + 5v = 11v at the input voltage.
So keep these in mind when you pick a transformer.
For 6v and 12v, I would pick a transformer that has two independent secondary windings of about 9v AC and if you want up to 10A, then you'd need one rated for at least 18v x 10A x 2 = about 300-350VA
If you want to output 6v , you connect only one winding to the bridge rectifier and capacitors , if you want 12v you connect both secondary windings.
So let's say we have 2 x 9v AC and 300VA transformer and you want 12v and 10A.
The 18v AC is RMS, which means when you convert it to DC using a bridge rectifier, you get a peak DC voltage of Vdc peak = sqrt(2) x Vac - 2 x Vdiode = 1.414 x 18 - 2 x 0.8 = ~24v DC
The maximum AC current is Iac = 300VA / 18v = 16.66 A and the max. DC current would be around 0.62 x 16.6 = 10.3A which does the job.
Now you need to make sure the minimum voltage always stays above around 13.5v so that regulators like LM1084 will output 12v even at 4-5A per regulator ... so now you figure out how much capacitance to use :
C = Current / [ 2 x mains frequency x ( Vpeak - Vmin) ] = 10A / 2 x 60Hz x (24 - 13.5) = 10 / 120 x 10.5 = 0.00793650 Farads so you'd need a minimum of 7936 uF capacitance to keep this minimum voltage above 13.5v .. so two 4700uF 35v rated capacitors in parallel should do the job. You can go with more capacitance but you'd only get the minimum voltage higher at lower currents which means the linear regulators will get hotter.
For 6v, you could connect the 2 secondary windings in parallel, which means you'll have 9v AC and up to about 20-25A of DC current available.
Still, 300va transformers are quite expensive, so you can see why people would prefer to use a switching regulators, because they're more efficient. But in your application (radios) they may not be a good idea.