Difficulty for a fireman to hold a hose which ejects large amounts of water at a high velocity A discussion question in my book is:

"Why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity."

Its answer is given as:

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

My doubt is does water flowing in a pipe push the pipe really backwards??
 A: Water flowing through a rigid nozzle creates a small backward force that is negligible compared to the forces of the overall hose.  Consider the difference in difficulty when the hose is stretched out in a straight line on the ground for its entire length.  Holding the nozzle down on the ground so that it does not leap around or push itself backwards is easy relative to moving the hose around and spraying down a burning house.
Additional forces exist behind the nozzle and inside of the hose: water flowing in one direction wants to remain flowing in that direction.  As the fireman moves around and curves the hose, he must continuously provide forces to accelerate the mass of water around any existing curves of the hose.
Those curves are constantly fluctuating, the water is fast moving and heavy, and so this is a hard task.
A: 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.
A: You are right, somewhat.
For a straight hose, the naive argument is that since the hose pushes the water forwards, the water must push back on the hose by Newton’s third law, resulting in a backwards reaction force. In terms of momentum, this argument may make more sense: if the initial momentum is $\vec p = 0$, then this must be conserved, meaning that there must be a backwards momentum when the water is going forwards.
However, the fallacy in such an argument is that we assume the firefighter to be part of the action-reaction pair. The true action-reaction pair is that between the water and the pump (and wall behind it). The pump will experience a non-negligible force acting on it after it releases water.
The real reason the firefighter will struggle is when the hose is not completely straight. Water will push on the pipes at places of curvature, and given its  large amounts of water and velocity, ($F = \frac{\text{d}p}{\text{d}t} = \frac{\text{d}m}{\text{d}t}v = \mu v$) the force at these bends will be very high and transmitted to the firefighter. So the firefighter will have to struggle to make the hose straight.
References: Vera, F., Rivera, R. & Núñez, C. Backward Reaction Force on a Fire Hose, Myth or Reality?. Fire Technol 51, 1023–1027 (2015). https://doi.org/10.1007/s10694-014-0430-5
A: In addition to the answer by @bovflux, note that the idea of a straight hose or column of water, pushing back against the pump/outlet, is false.
From Wikipedia:
"The column will remain straight for loads less than the critical load. The critical load is the greatest load that will not cause lateral deflection (buckling). For loads greater than the critical load, the column will deflect laterally."
https://en.wikipedia.org/wiki/Euler%27s_critical_load
What this means is that the hose will always buckle, and a function of a three-man team holding the end of the hose is to prevent it whipping around.
A: 
does water flowing in a pipe push the pipe really backwards??

Water flowing steadily in a straight pipe does not push the pipe backward. Water flowing through a nozzle or a curved pipe does. The difference is the acceleration of the fluid.
A: The easiest way to visualize this might be in a practical example. If you've ever seen a water rocket, you seen that the water really does push against the vessel it's contained in.
In your firehose scenario, the fluid (water) is actually pressing against the walls of the hose throughout the entire hose. Ultimately at the end of the hose opposite the nozzle is one of two things (I'm not an expert on fire trucks, but I've got a pretty solid background in fluid mechanics): there is either a pump or a pressurized container (which, in turn, was pressurized by a pump). The water in the hose is pressing back against that pump or pressurized container (which is in turn pushing against the fire truck, which is in turn pushing against the ground via friction), and the force with which it is pushing is roughly equivalent to the force at which it is being expelled from the nozzle (there is also friction to overcome in the hose and nozzle).
When the nozzle is open, the firemen need to hold the hose rigid because without them, the pressure of the expelled water moves the hose. You've probably seen a flailing high pressure hose at some point, whether it was water, air, or some other fluid. If the hose itself was rigid (like a pipe), the hose (pipe) would resist that motion (by providing equal and opposite force), and firemen wouldn't be needed.
A: 
As shown in the figure, this is a convergent nozzle, and water flows out of the nozzle from the right. Its left side is connected to the hose. The force of the water upstream of the nozzle to the water in the nozzle is $F$, and the reaction force of $F$ is $F'$. The force of the shell of the nozzle to the water in the nozzle is $f$. The reaction force of $f$ is $f'$. Although, the axial component of $f'$ pushes the nozzle along the axial direction. However, $F'$ will lead to the bending of the hose connected with the nozzle. And keep swinging. This is the recoil force of the convergent nozzle.

As shown in the figure, we ignore the viscosity of the water and apply a resistance $δ$ to the water by the convergent nozzle. We let $β=F-δ$，$β'=-β$， that $β'$ It is the recoil force, which acts on the water and points upstream, bending the water column.
$δ$ Is the axial component of $f$.
