Would 200 lb of water thats 150 ft high lift more than 200 lb due to the weight + the extra pressure of the fluid being so high up? 
I'm trying to understand Pascals principle in more detail so I made a drawing with a question to find out a segment of the differences between pressure and weight.
As demonstrated in the project "Pascals Barrel" the deeper depth of water the greater pressure on the lid of the barrel below would be. Here I've drawn a photo replacing the barrel to a simple type of hydraulic lift so that I can figure how much weight could be lifted and how much of a distance that it would be lifted upwards. It would help to figure out its equal equilibrium too.
To keep it simple please disregard any extra fluid or pipe in the drawing and just take the data that's provided below.
2" pipe that's 150 ft high = (24 gallons or 200 lb of water).
Say this pipe is stagnant water. Because pressure and weight are of different things and assuming the fluid is now adding both the pressure and weight onto the piston below (at sea level) how much weight would that 24 gallons of water lift and how much of a distance would it be lifted? And or be at equilibrium? 
That's 200 lb of fluid + the pressure it being in a skinny pipe reaching 150 feet in height. 
 A: The water pressure exerted by the fluid already includes the weight of the water. In fact, the pressure is caused by the fluid's weight.
Therefore, the force exerted by the water onto the piston is just the water pressure at the piston multiplied by the area of the piston. The water pressure is given by 
$$p=\rho g\Delta h=\frac{mg\Delta h}{A\Delta h}=\frac{mg}{A}$$
where $\rho$ is the density of water, $g$ is the acceleration due to gravity, $\Delta h$ is the difference in height between the top of the water and the piston, $m$ the mass of the water contained in the height $\Delta h$, i.e. the water in the left tube between the top of the tube and the height of the piston (anything effects from water in the bottom of the U shape cancels out), and $A$ is the cross-sectional area of the pipe. 
The first expression might be what you are more familiar with for the pressure difference between two heights in a fluid, but as you can see, it just amounts to the weight of the fluid in the relevant part of the pipe. i.e. $F=pA=mg$.
If we were to push on the left side with an additional force $f$, we would still just get that force out on the right, the pressure would just be $(mg+f)/A$. Instead, this type of system becomes useful when the areas on either side of the tube are different. This is where you get "more bang for your buck" in terms of pressure applied on the left and and pressure obtained on the right. I will leave this for you to explore.
