What is the total power consumption of a flipflop?
Assume that \$C_{PD}\$ is charged each time the flip-flop is clocked and assume Vcc = 5 volts. Also assume that \$C_{PD}\$ is discharged at the same rate (just for convenience in this example I'm giving). Different logic configurations will have different results so I'm just giving an example.

Energy put into \$C_{PD}\$ when charged = \$C_{PD}V_{cc}^2/2\$ and this is discharged (turned to heat) by the discharge process cyclically at the operating frequency F. This then tells us the power consumed is \$FC_{PD}V_{cc}^2/2\$. If we divide power by the supply voltage \$V_{cc}\$ we get the average current lost by charging up and discharging this capacitor, \$FC_{PD}V_{cc}/2\$ .

I note that from the blue box in the question, the dynamic average current is stated as \$FCV_{cc}\$ so maybe this can be justified elsewhere in the data sheet but, for the physics of charging and discharging a capacitor at frequency F, my answer is correct.


  How can I get the load current?


You say in your question that the load is another identical flip-flop hence, you can perform the same maths as above to calculate average load current. I don't know what output frequency is being produced but presumably you do so. do the simple math!

If you also have a resistive load that is switched on and off then the average current for that is simple ohms law multipled by duty cycle.

\$I_{cc}\$ is listed in the data sheet on page 3.