How can a guy calculate the second collector current from this circuit?
Using a Darlington transistor Vce(sat is higher)

Then using two with isolated collectors   the 1st does not saturate and the 2nd does which is why you concluded it was independent of hFE. ( Ic=Vcc/Rc)


But a Darlington transistor ties both collectors together, so there is a Vbe drop added to the output.



But in your 1 Ohm load case it is operating in linear mode limited by total Hfe



Note above uses hFE=200 for each   3A:15mA:75uA

They don't teach you this at schoole or show in datasheets

When a transistor saturates hFE reduces to ~ 10 to 20% of hFE max at Vce=Vce(sat)
What they guarantee in datasheets is Vce(sat) @ Ic for Ic/Ib=10 to 20 per transistor.

So what Rb is needed to saturate 1 Ohm @ Vcc=9V near 8A \$Ib=\dfrac{8A}{h_{FE1}*(h_{FE}*10\text{%})}\$  =2mA so Rb= (9V-1.4V)/2mA = 3.8kohm

Other stuff, neg. feedback

Now see negative feedback situation.