Choosing a resistor for balancing lead acid batteries
The ideal way to transfer charge between two batteries with substantial voltage differences is with a buck converter, with efficiency able to be in excess of 90%.
However, at low voltage differentials the use of a resistor may be more efficient!
Power input = Vin x I. Power Output = Vout x I.
Efficiency = Pout/Pin = Vout/Vin.
For eg 12.5V in and 12V out, efficiency = 12/12.5 = 96% !  

Use of a resistor is better than "just a thick wire" for safety purposes but regardless of what form of resistor you use you will dissipate energy as heat in the process.
The resistor value could be set based on known conditions, or a worst case value could be used.  

Rminimum = (Vsource - Vtarget) / Imax_allowed.  

If you have two batteries with say 12.8V (about fully charged in rest state) and 12V terminal voltages and can tolerate say 10A current then R = V/I = (12.8-12)/10 = 0.08 Ohms.  

ie enough "thinnish" wire may suffice.
Some prior experiment (or measurement) could allow you to size a wire that limits current to some maximum value and does not get too hot (or glow in the dark) in the process.
Something around a few 10ths of an Ohm all up (wire, clip connections, contact resistance) would probably suffice.

The danger is in using eg jumper leads with batteries of very different charge levels.
Many 10's of amps could flow and, very worst case - but unlikely, 100 + A.

One possibility is a low voltage light bulb - say a 3V bulb (although where you'd get one of adequate current rating is hard to imagine). As current increases bulb resistance will rise and maintain "a degree of" current limiting.