Is it not possible to calculate this? An empty but healthy battery will have maybe 8 or 10 V. An alternator will usually produce 14 V to charge the battery.
So there is a 4 to 6 V potential.
The typical battery internal resistance is 0.02 ohm. (I looked this up on google)
Using the formula U=R/I and solving for current I=U/R.
And using the above values I get
Imin = 4/0.02 = 200 A
Imax = 6/0.02 = 300 A
I am assuming that the resistance in the cables is 0. This is not true, the resistance in the wires to this battery will drop the current some.
But this would be the instantaneous current, as the battery charges the voltage difference decreases, and thus the charging current decreases.
Also the alternator might not be able to support this high current, so the voltage over the primary battery and alternator will fall when connected to the empty secondary battery.
It was mentioned that the alternator can give 120A. I assume that this is at 4000 rpm or something like that engine speed, not at idle.
So if the empty battery is connected to the primary battery and alternator, and the motor is being revved so the alternator can produce 120A, then nearly all the 120A will charge the secondary battery.
But at this rate of charge you will fully charge your battery in probably around than 1 hour. (180 Ah rating for battery as stated earlier). And you will likely only pull the full 120 A the first minute or so, after that the voltage difference decreases and the charging current falls with it. And it is likely only the full 120A if you are revving the engine at the optimal speed for the alternator to produce that current.
I think this becomes an issue of duty cycle and heating of the wire as well.
Can you measure what wire size is coming out from the alternator? Then use the same gauge wire, maybe a tad larger if you want to feel safe, to connect your secondary battery?
Probably the people at Bosch or whoever built your alternator have figured out what wire gauge is suitable.
Ok, but what happens if just the two batteries are connected together in parallel? One empty, one full.
I found this link.
https://www.powerstream.com/voltage-sag-during-engine-starting.htmI read it as the battery can provide around 150A, but then the voltage drops to around 10V. A healthy battery is 12.8 V when fully charged, so 2.8 V drop. I think that this current can only be this high for the first couple of seconds. Then as the voltage difference decreases, the current also decreases. I doubt that the same gauge as jumper cables would be necessary, but probably more than AWG32

.
I found an interesting article that might help
https://www.itacanet.org/eng/elec/battery/battery.pdfI was writing as I was thinking. I hope this helps.