Dual optocouplers often have the second pair of inputs reserved. This is also the case for the FOD053L; the symbol is wrong, you have to exchange pins 3 and 4 in the schematic.
There is no single CTR value because the actual CTR varies greatly due to manufacturing tolerances. (And temperature, and forward current.)
To compute the required resistors for the input LEDs, go backwards from the outputs: with the 1K pullup resistors, the outputs need to sink about 3.3 mA. With the worst-case CTR of 15 %, the LEDs need a current of about 3.3mA/15% = 22 mA. With a worst-case voltage drop of 1.7 V over the LED, the input resistor needs to drop 12V-1.7V = 10.3 V, so at a current of about 22 mA, the resistor must not be larger than about 10.3V/22mA = 468 Ω.
Please note that 22 mA is dangerously near the absolute maximum, and this maximum is likely to be exceeded with a higher input voltage. You should consider increasing the pullup resistors; e.g., doubling them results in half the required LED current.
(Increasing the pullup resistor makes the output switch off slower, as shown in figure 10 of the FOD053L datasheet. However, this is probably not a problem in your case.)
The FOD053L has a lousy CTR because it is a high-speed optocoupler. If your signals aren't faster than about 10 kHz, consider using a plain phototransistor optocoupler in a small package, such as the PC3H7 or one of its clones (Fairchild's is the HMHA2801).
