Current measurement in 12V LED circuit
Can't tell the model number from your picture, but check the isolation or common mode range of your device.  I'm betting it can't handle the voltage difference between negative inputs.  Two choices: 
1. Arrange the circuit using one channel for each side of the shunt, with all three channels sharing a common ground, and subtract shunt low from shunt high to get your current.
2. Put your shunt on the return side of the battery.  This is the less attractive choice for accuracy because of the IR losses in the power wire, if there is a common ground in the logger.  If there is some isolation between the low side of the channels, you will still want a separate wire from the low side of the shunt to the logger to avoid this effect.

The shunt's current rating value depends on your logger resolution and your expected load times.  At 30 A, you will have 15 mv, so you will dissipate 450 mW in the shunt.  Lower current shunts have progressively higher resistance so you would dissipate more power when at 30 A.  If you are going to be at 30 A continuous, I would agree that you should get a 50A shunt, but if you only expect short excursions to 30 A you can go lower.  Go with the lowest power rating you can (highest resistance).  You might try using the common ground approach and set the  logger channel to its highest gain (lowest input voltage range).  You probably need a resolution of at least 1% of your shunt voltage (500 uV) to read the current and get usable information.  If the channels all have the same range and resolution, you need 16 bits of resolution to measure 12 V and the shunt with the same gain. 

Your part number has accuracy of +/- 0.1% FSR; with 30 volt input that is +/- 30mV, not enough for your shunt to be read accurately. It also has a common mode input range of 0 to 2.5 volt. You will have to put an amplifier on your shunt. Luckily you have 12VDC available to run an op amp!