Below, we see the accordion amplifier's topology, two output tubes, one atop the other, with an output transformer wedged in between. Yes, it does look push-pull in nature; yet, it works strictly in a single-ended fashion. And yes, it does look hopelessly asymmetrical, with what looks like a cathode follower on the top and a common-cathode amplifier below; yet, perfect symmetrical operation can be observed.

   In the accordion amplifier, like the parallel single-ended amplifier, both tubes conduct equally and in unison. When both output tubes increase in conduction, the top triode pulls its cathode connection to the primary up towards the B+ voltage and the bottom triode pulls its plate connection to the primary down towards ground. When both tubes decrease their conduction, the top triode allows its cathode to swing down towards ground and the bottom triode allows its plate to swing up towards the B+ voltage. (Yes, they do meet and ultimately pass each other, the top triode's cathode being at a lower voltage than the bottom triode's plate voltage.)
    If the phase of the input signal to one of the parallel or accordion amplifier's output tubes is reversed, the amplifier ceases to amplify, as the output transformer does not see a change in current flow; and it is the change in current flow  through the primary that drives the secondary. Once again but in greater detail, since the output transformer relays the delta (the difference) in current conduction through its primary, if the primary sees a constant DC current flow, then the secondary will remain idle. In these examples, the net conduction can never break away from the idle amount, as the output tubes conduct in anti-phase. Wait a minute: one output tube sees an increasing grid voltage, while the other sees a decreasing grid voltage, so how can the current not change?
    Let's consider the parallel amplifier first. If the phase of the input signal to one of the output tubes is reversed, then one tube will increase its current conduction while the other's decreases. The sum will equal the idle current, as one tube's increase in conduction is matched by the other tube's decrease in conduction. For example, +10 mA added to -10 mA equals 0 mA. Of course, real tubes are not perfectly linear, so there will be some small change in net current flow equal to the imperfection of tubes and their relative match to each other. (In fact, this might be an excellent test of the degree of matching and non-linearity.)   

     Before going into the details, let's look into what can be seen in the above schematic. We see that both tubes are presented with an in-phase input signal, compelling an in-phase current conduction, thus eliminating the need for a phase splitter because of the single-ended functioning. Compare this to a push-pull output stage, defined by the output devices working in anti-phase conduction to each other, i.e. as one conducts more, the other conducts less.

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