I will focus on just a little bit of one of your questions - the relationship between compressibility, density and pressure - and per my comment, recommend that you narrow down the scope of your question.
As you know, in a gas we experience "pressure" because molecules hit the walls of the containing vessel. When I double the number of molecules in the same volume at the same temperature, I double the number of collisions (each imparting on average the same momentum) and thus double the pressure - this is the familiar ideal gas law.
Now when the size of the molecules becomes a sizable fraction of the volume, the rate of collisions goes up. Imagine a pingpong ball between two walls. If the distance between the walls is large compared to the size of the ball, the time for a round trip is inversely proportional to the size of the ball; but as the distance approaches the size of the ball, the rate of collisions goes up rapidly.
I think a similar thing happens with "nearly incompressible" liquids: there is a small amount of space between the molecules, but they are permanently bumping into each other and into the walls of the vessel. As you increase the pressure, they bounce more frequently as they have less far to travel before they collide with another molecule (or the wall).
All this is still treating the liquid like a non-ideal gas. In reality, not only do you have the finite size of the molecules, but also attractive forces between them. Both these things make the picture a bit more complex than I sketched. But I would say that the above reasoning nonetheless applies (with caveats).
As for the experiment you described with stoppers on the inside or outside - there are other things going on there as you go from the static (no flow) to the dynamic (flow) situation - the water needs to accelerate before it will flow out at a certain velocity. But I think all that should be the subject of another question.
Let me start off with something that should be obvious but I feel the huge need to say it -- I am not trained in the biomedical field and what I put in my answer should not be construed as definitive and ready-for-use in actual patients. I will do my best and present what I think is correct, but I am not a doctor and don't even pretend to play one on TV.
Okay, that out of the way, this is an interesting problem. You mention using Bernoulli's equation and that would be really handy to use here if we have all the information. For completeness, the equation says that for an incompressible (which blood is close enough to one, so we're good there), isentropic (so no heat exchange into/out of the system, no irreversible losses... blood isn't so great here) fluid, the total pressure is conserved:
$$ P_0 = P_s + \frac{1}{2}\rho V^2$$
Let's look at what we would need to know in order to apply this. The right atrium, RA, is basically a stagnation chamber -- blood is at rest at a particular pressure which causes it to flow out of the chamber into the catheter and through the tubing. We don't know what this pressure is, but the pressure is $P_0$. We have a pressure probe at P3 that is probably measuring the static pressure, $P_s$, so we have that number. We also have the flow rate of blood, which is going to have units of mass per unit area per second. Let's call the flow rate $\dot{f}$. The flow rate is related to the velocity of the fluid by $V = \dot{f}/\rho$. This should give us everything we need to figure out the total pressure in the RV:
$$ P_{RA} = P3 + \frac{1}{2} \rho \frac{\dot{f}^2}{\rho^2} = P3 + \frac{1}{2} \frac{\dot{f}^2}{\rho}$$
Now... this will be okay, provided the assumptions we used are valid. The incompressible assumption is okay, I'm not too worried about that. The isentropic assumption though, that one is more concerning. It may not be big enough to matter in practice depending on the terms involved, but it could change the answer. Whether the change is large enough to be important, I don't know. Anyway, here goes.
There are two assumptions for an isentropic flow. Adiabatic and reversible. Adiabatic means we are not adding nor removing energy from the system. But in this case, we are. There is heat exchange with the environment between the tube and air. I don't know how much heat is lost from the blood once it is in the tube. We also add energy because of gravity. This one, we can account for though and I will modify the equation below to do so.
For reversibility, that means there can be no losses in the flow. No shock waves, which again I think is fine here, but also no viscous forces. Blood is viscous. In fact, it's a non-Newtonian fluid, which makes it harder to model. Fortunately, it is shear-thinning and so the viscosity decreases as more force is applied. This may mean the viscous losses are small, but I just can't put any justification behind that statement.
Okay, so back to gravity. The equation is modified:
$$ P_{RA} + \rho g h_{RA} = P3 + \frac{1}{2} \rho V_{P3}^2 + \rho g h_{P3} $$
where $h_i$ is the height of the measurement location(s). You can then substitute in everything you know (making sure your units are consistent of course) and find $P_0$.
Back to the unknown losses. We might be able to come up with an estimate for them using the information at P3 and P4. We know:
$$ P_{0,P4} = P4 + \frac{1}{2} \rho V_{P4}^2 + \rho g h_{P4} $$
and we also know that:
$$ P_{0,P3} = P3 + \frac{1}{2} \rho V_{P3}^2 + \rho g h_{P3} $$
We can then find the change in total pressure between those two points. Ideally, this would be zero/very small. Then we know that Bernoulli's equation will hold quite well. But in reality, it won't be zero. Hopefully it will still be small though. Once we compute the difference between these two points, $\Delta P_0 = P_{0,P4} - P_{0,P3}$, we can find how much loss there is per unit length:
$$ \frac{\Delta P_0}{\Delta s} = \frac{\Delta P_0}{s_{P4}-s_{P3}}$$
where $s$ is the coordinate along the tube axis (to account for it turning, etc.. If it were completely vertical and straight, this would just be the difference in height between the two points).
We can then make an assumption that the losses between P4 and P3 per unit length are the same as the losses between RV and P4 per unit length. This is probably not true in reality, because heat loss through the catheter inside the body will be different from heat losses from the tube to the air, and the viscous losses will be different also, but we don't have the info here to measure that. So we'll just lump that all into a coefficient $\epsilon$. This will give us the final, complete equation as:
$$ P_{RA} = P3 + \frac{1}{2} \frac{\dot{f}_{P3}^2}{\rho} + \rho g (h_{P3}-h_{RA}) + \epsilon \frac{\Delta P_0}{\Delta s}(s_{RA}-s_{P4}) $$
where $\epsilon$ is going to hopefully be something we can figure out (and heck, we can design some experiments to compute it -- but that's another question), but until you can figure out the real value of it, we'll have to roll with taking $\epsilon = 1$ for now.
Best Answer
When a pipe is bent the outside curve - what would be the longest path through the curve - has the highest pressure and the lowest speed. The inside curve - the shortest path through the curve - has the lowest pressure and the highest speed. In short, when the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always higher than the pressure on the inside of the bend.
The pressure would be lower on the lower right inside point of the bend.
I could no view that attached image; however, I hope that helped.