The answer that it is the flow at the nozzle is all that matters is partially correct. The nozzle has a constriction, so that the water flows at a higher speed out of the nozzle than through the hose. That means it is accelerating at the point of the nozzle, and therefore you can infer a force.
However, water flowing through the hose also undergoes an acceleration if the hose is curved. Let's say for simplicity that everything takes place in a horizontal plane. Also assume WLOG that the fireman points the hose at zero degrees. Then there is only one variable you need to consider for the hose geometry, the angle the hose exits the source (hydrant, engine, etc.), theta. That might be theta = 30 degrees, if the hose makes only a slight curve, or theta = 90 degrees, if it makes a right turn, or theta = 135 degrees, if the hose makes almost a complete U-turn. The exact path that the hose takes is irrelevant to the solution. It could follow an S shape or go around a complete circle three times. The only question is which way the water enters and which way the water exits.
Now the water in the hose will be moving at a different speed than that coming from the nozzle. Let's define the diameter of the hose to be D > d. The speed of the water in the hose is |v|d/D. The impulse is delta m times the difference in velocity at the source and the nozzle. The speed is the same, but the velocity is different. It has x and y components:
I.x = delta m (|v|(d/D) - |v|(d/D) cos theta)
I.y = delta m (-|v|(d/D) sin theta)
The mass flow rate is rho pi (d/2)^2 |v|; we'll assume rho = 1. The force that needs to be exerted to keep the hose stable is
F.x = pi (d/2)^2 |v| |v|(d/D) (1 - cos theta)
F.y = -pi (d/2)^2 |v| |v|(d/D) sin theta
However that doesn't get you directly to your solution. That force is applied at two points: the source and the nozzle. Those two components have magnitude equal to the tension of the hose, but different direction. The directions don't have to be the same as the exit direction of the source and the aim of the nozzle; the hose can "kink" at these points.
At this point I will leave the rest of the solution unfinished. I believe you can proceed by assuming the hose will take up the shape of a circular arc using its full length. The angles of the tension forces thus depend on the length of the hose and the relative position of the fireman to the source. You can solve for those angles, then for the tension.
Incidentally, the answer to your original problem should really be found by F = dm/dt (|v| - |v|d/D). The delta m (|v|d/D) component of momentum is taken up by the water source. To find the answer for the force the fireman must exert when the hose curves, add that solution and the tension as vectors.