Electronic – Braking with an electric motor

Those wires form coils, so are long. Every bit of wire has some resistance, and all those bits of wire end to end result in a significant enough resistance to not look like a "short".

These wires shorted across a voltage source is exactly where the stall current of the motor comes from. It is simply the voltage applied to the coils divided by their resistance.

When the motor is running, then another effect is present. The motor actually acts like a generator so that spinning in the forward direction cause a voltage to be generated across the coils. This voltage opposes that applied by the external power source. The current thru the motor is therefore the power voltage minus this reverse EMF produced by the motor acting as a generator, and that result divided by the coil resistance. The faster the motor spins, the less current, because a higher back EMF is subtracted from the driving voltage.

This back EMF effect also limits the top speed of the motor. At some speed, the back EMF generated internally cancels the applied voltage, and nothing is left driving the motor. Of course it wouldn't spin at that speed since nothing is driving it anymore, but it works at a little lower speed if nothing is loading the motor.

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Figure 1. OP's circuit redraw with positive voltages on top and negative on bottom.

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Figure 2. Using snubber diodes. In this configuration the maximum voltage that can appear across the contacts is 12 V + 2 x 0.7 V = 13.4 V.

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Figure 3. Back to back Zener diodes. Zener diode voltage should be a few volts above supply voltage to prevent turn on or high leakage around the knee voltage.

The solution of Figure 3 is probably the most elegant as it uses only two components which can be placed at the motor and requires no additional wiring.

Where should the snubber diode go to protect RL1? (Between 5 & 4 would flow current back into +12v which seems like a bad idea. Between 1 and 2 wouldn't work because when the motor is running in one of the two directions the diode would short. I have reviewed lots of diagrams and descriptions but can't get my head around the correct location. )

You are correct that there can be a problem. Since you are trying to protect the contacts the best place to do the protection is at the motor (but see Figure 4, etc.).

Am I correct that the objective is to use the diode voltage drop (.6v) to consume the inductive kickback as current cycles through a circuit containing the diode.

Yes, but 0.6 V isn't a magic number. You just need to keep it low enough to prevent arcing. Figure 3 shows a possible solution. In either case one Zener diode is forward biased (0.7 V drop) and the other is reverse biased. The reverse biased one will limit maximum voltage to the sum of the 0.7 V + \$ V_{Zener} \$. This will work in either polarity or if RLY2 is switched while live.

Is inductive kickback a risk to RLY2? I know it is to RLY1.

Not if it doesn't switch the live load.

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Figure 4. If RLY1 does all the load switching and RLY2 only switches when power is off then a diode placed as shown will suffice.

As @user28910 points out, Figure 4 will suffice if sequencing can be guaranteed.