There's a pretty good website here. Here's the first picture of the equivalent circuit of the whole motor: -
The section in the OP's question is the secondary side of the picture i.e. the rotor. Resistor \$R_R\$ represents the real mechanical output power of the motor and \$jX_R\$ represents the impedance of the rotor winding.
The transformer is there because the rotor is coupled to the stator by transformer action. This diagram is only one-third of the full picture of a 3-phase induction motor.
The website goes through the relatively simple theory of converting the circuit to its more usable form: -
Here, the term "s" is used on the rotor output (represented by \$R_2\$ now) to properly relate output power to the slip of the rotor compared to synchronous machines that do not slip.
20 kW is the rated mechanical output power. The actual operating power depends on the load. Subtract the core loss and the stator copper loss to get the air gap power (the power transferred from the stator to the rotor across the air gap). Subtract the rotor copper loss, friction and windage (air drag) losses from the air gap power to get the mechanical power delivered to the load in watts (746 W = 1 Hp).
You are correct; n < ns indicates operation as a motor. If n > ns, you would add the losses to the electrical power instead of subtracting and the electrical side figures would be the output and the mechanical side would be the input.
Best Answer
The power in R2/s is the sum of rotor copper losses plus electrical power converted to mechanical power. To look at both the losses and the mechanical power, you can divide R2 into to two parts, R2 and R2(1-s)/s. The rotor current is proportional to E2xR2/s, so the rotor current and the losses in the rotor are directly proportional to slip. The mechanical power developed in the rotor is also proportional to slip.