It will output a pulsed current with 4mA low current and 18mA high current (+- 1mA).
The frequency will determine the velocity as can be seen in the graph in the datasheet:
The high pulse will have a duration of 100µs +-20µs, so it's more like a PWM output than a rectangular frequency.
To answer some of the follow up questions in the comments:
To read the signals you would probably use a resistor in the output to convert the current output to a voltage output, which can then be read by some input hardware. The total resistance is limited by the sensor and is given as 600\$\Omega\$, which includes the wire resistance.
Why would someone use a current pulse output?
Because an anemometer is typically placed in a remote location the wire length to the sensor can introduce a significant resistance. If you would use a voltage output, the resistance will distort the signals amplitude, which might be okay for a pulse output. But a current output is also more robust against capacitive coupling.
How can you get 0V to 8V for your pulse input?
A simple solution might be this (I'm not sure this will actually work, there were some glitches in the simulation, maybe someone can comment on that):
simulate this circuit – Schematic created using CircuitLab
But you should be aware, that the shunt resistor is quite close to the maximum allowed resistance in the output of the sensor. You also loose the information if the sensor is dead as you won't have the live zero signal present behind the capacitor.