Push-pull amp in class A what it mean
They both must have opposite direction and must differ from each other and the 0 must be on load.

What makes you think that the output currents of the NPN and PNP transistor have a different direction? You should be careful with the direction of currents and also separate the direction of a DC current (bias current) and the direction of a change in current (due to a small input signal) which we often call AC.

Assume that we have a small but positive input signal, meaning the voltage rises. What will happen:


The \$V_{BE}\$ of the NPN will increase
The \$I_C\$ and thus \$I_E\$ of the NPN will increase
The \$V_{BE}\$ of the PNP will decrease
The \$I_C\$ and thus \$I_E\$ of the PNP will decrease


So at the node "Probe1" there's a current increase coming "from above" while at the same time there's also a current decrease coming "from below".

And the current law tells us that the current into load R3 will be:

\$I_{load} = I_{above} - I_{below} = I_{NPN} - I_{PNP}\$

So the currents do not have cancel but actually add up.

But there's a minus sign so they subtract?

Yes but note how \$I_{NPN}\$ increases yet how at the same time \$I_{PNP}\$ decreases. So those changes do add up.

When the NPN supplies more current it is also "helped" by the PNP as the PNP will draw less current.

It is like the NPN is "pushing" more while at the same time the PNP is "pulling" less. So both NPN and PNP contribute to the change in the current through the load.