A force is applied to the nozzle because it accelerates the water. It's like a rocket.
The pipe (on the right) has a cross section area S' and water is going at velocity v'. Thus at the nozzle intake we have a mass flow rate of rho.S'.v' with rho being the density or mass per unit volume.
The nozzle is conical, and has a smaller exit cross section area S. At the exit, water goes at velocity v. The mass flow rate is the same at the entrance and exit of the nozzle, therefore:
rho.S'.v' = rho.S.v
From this, the exit velocity is the intake velocity multiplied by the ratio of the intake and exit cross sections. Basically, squeezing water through a small hole needs it to go faster than through a big hole.
This acceleration implies a force, which is the mass flow rate multiplied by the difference in velocity between intake and exit.
In addition, water at the exit is at pressure P, but water right after the exit is at atmospheric pressure Patm. This pressure exerts a force on the nozzle, which is S.(P-Patm).
The sum of the two gives exactly the same equation as a rocket nozzle. The shapes are different, because the rocket nozzle is meant to work with gas, not liquid.
The water in the pipe behind the fireman exerts a force S'P' on the nozzle in the other direction, trying to pop it off the pipe. But this force is counteracted by the fitting holding the hose to the pipe. This force applies to the pump at the end, trying to stretch the whole pipe. It does not apply to the fireman.
This is made experimentally evident by playing with a water hose with or without a sprinkler on the end. When the hose has no sprinkler, no nozzle, and no fitting that would act like a nozzle, the reaction force is much lower than with a nozzle fitted, even though the mass flow rate is the same.
This does not account for bends in the pipe. If there are bends, then the water has to receive a centripetal force in order to turn, which means the bend in the pipe receives a centrifugal force. Basically the water will try to straighten the pipe.