So let's assume a few things.  If the batteries differ by one volt then you
have one volt with 50A capability applied across a five foot 14AWG wire.
For a greater voltage difference you will have more available current say up
to 150A.

I seem to recall that the headlights are wired with 14AWG.  They draw ~5A
and the wiring does fine. BUT the current is limited by the filament which
has essentially no resistance when cold but instantaneously rises when
current is applied..

The resistance of five feet of 14AWG is the question and if I had Audels or
Terman handy it would be settled.

The simplest way to test this is with a variable power resistor and a 12V
battery.  With the wire in series with the resistor across the battery with
it set to 500 ohms or so.  Measure across the resistor. The voltage should
be very close to the battery voltage.  If you know the resistance then you
know the current by Ohm's' Law.   Repeat with lower and lower values of
resistance until you calculate 15A current based on the voltage across and
resistance of the resistor.  ( you could use a calibrated ammeter of course)

The battery voltage less the voltage across the resistor is the voltage drop
in the 14AWG wire for a 15A current.  I suspect that voltage to be less than
one volt.

Now repeat for a two volt drop across the WIRE.

Phlogistics theory will probably apply under this set of test conditions.

For a 10% voltage drop (say 1.3V) across 10 feet of wire 14AWG is RATED for
30-40A.

The ampacity for a wire is expressed as follows:  CM=K*I*L/E

Where CM is conductor area in circular mils (from table)
K=10.75
I =current in amps
L=round trip length in feet
E=Voltage drop across L in Volts

Minimum circular mils for 14AWG is ~3700.

So a nomogram of I vs. E for a 5 foot length is left as an excercise for the
student.

E of course will dicatate the maxmimum voltage difference between the
batteries to stay within the 15A limit.

pensioner