When the bottom tube conducts more current than the top tube, the difference in current flows through the bottom tube's plate into resistor Rak and then into load resistance.
    Wait a minute: doesn't current only flow from negative to positive? How can the current flow from a very positive cathode, 100 volts let's say, into load resistance at ground potential? The answer lies in the use of a coupling capacitor that connects cathode to the load. The capacitor (being very large in value) functions like a battery; in fact, it could be replaced by a high voltage battery. The capacitor maintains an almost fixed voltage across its leads, so when the bottom triode conducts more, it will pull the capacitor down negatively, making the connection with the load resistor become negative relative to ground, which in turn will allow current to flow up through the load resistor into ground. If you are still unclear how this could be, imagine that the bottom triode conducts so fiercely that it becomes effectively a dead short. Now, the voltage charge stored in the capacitor is still present and it will begin to discharge through the load resistance (and resistor Rak) until it is completely discharged. Conversely, if the top triode conducts so fiercely that it becomes effectively a dead short, the capacitor will begin to charge up through the load resistance from ground until the charge across the capacitor equals the B+ voltage. 

Is Balance Possible?
   Normally, the prime consideration in designing any push-pull amplifier is balance. Balanced drive voltages that yield  balanced current swings make a clean push-pull amplifier. Matched tubes, equal cathode-to-plate voltages, equal idle currents--all are needed to ensure balance. But this circuit does not appear to be balanced in structure. Why is resistor Rak value not strictly specified? Why does the bottom tube have to work into Rak and the load, but the top tube directly works into the load? Should the bottom triode's cathode resistor be bypassed or not? Put more generally, can this amplifier be balanced at all?

   Yes, this circuit can be balanced by careful selection of part values and operating points for a given load impedance. Unlike the White Cathode Follower, the limitation to the SRPP is that it can only be optimized for one specific load impedance. Of course, if the circuit is optimized for a load of 10k, 15k will not totally ruin the amplifier's performance. But 600 ohms or 1M will prevent the optimal use of this circuit. Now this situation is not unusual, as most power amplifiers are designed with a fixed load impedance in mind in order to perform to their design standards. Still, this circuit is more sensitive to mismatching, as it does not benefit from the degenerative feedback and the loop feedback that keep the White Cathode Follower's output in line with its input signal. Furthermore, the SRPP uses the current delivered into the load impedance to define the drive voltage for the top tube; wrong load impedance equals wrong drive voltage and unbalanced operation.

    If the load resistance is zero ohms, then the difference in current flows indirectly through the coupling capacitor into the ground.