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.
The problem is that you're modeling the liquid like an ideal gas, whose molecules independently bounce off the ball, but liquids are characterized by strong interactions at short distances.
A better (but still inaccurate) model would be to treat the liquid like a solid locally, i.e. imagine each of the liquid molecules connected in a chain by springs. An increase in pressure means that the springs are compressed more and more, so they push outward onto your object more and more.
In terms of your variables, we should have $F \sim k \Delta x$, not $F \sim 2mv/\Delta t$. In this model, pressure can be transmitted from molecules far away, just like tension is transmitted through a rope.
Best Answer
I can think of a few effects in play:
First, when there are bubbles in the line, the "effective density" of the liquid will be lower - so if there is a drop $h$ from the infusion pump to the level of the liquid in the reservoir, there will be a pressure difference between the pump and the atmosphere of $\rho g h$. If $\rho$ drops, so does the pressure "pull" from the liquid, meaning the pump has to work harder.
Second, there may be surface tension effects. If you are trying to push a bubble into the liquid, the curvature of the air bubble and the surface tension of the liquid will cause an effective increase in pressure (in the bubble, and thus in the upstream system). The smaller the diameter of the bubbles, the bigger this effect will be (pressure difference is given by
$$\Delta P = {\rm{\frac{force}{area}}} = \frac{2\pi r \sigma}{\pi r^2} = \frac{2\sigma}{r}$$
Where $\sigma$ is the surface tension - which creates a force per unit length along the circumference of the bubble, and that force is countered by the pressure that works across the projected area of the bubble.
There might be other surface effects inside the tube as surface tension will interact with the air/liquid interfaces; in principle, these things ought to cancel (where one liquid/air surface might create "push", the next air/liquid interface might create "pull") but that's not necessarily true.
I suspect that surface tension is the main culprit - but without seeing the exact setup that is hard to confirm.