First a brief summary on what Bernoulli principle and the corresponding equation state in the simple case of air at equilibrium. As you mentioned, Bernoulli's effect tells you that energy is conserved along a streamline. Thus the sum of kinetic energy, potential energy and internal energy remains constant and an increase in the speed of the fluid – implying an increase in both its dynamic pressure $\rho v^2/2$ and kinetic energy – occurs with a simultaneous decrease in (the sum of) its static pressure, potential energy and internal energy.
Another simple view is the following: a pressure gradient between two points on a streamline causes a net force, a 'push' from the higher pressure area to the lower pressure one. Thus, by Newton's 2nd Law, acceleration of air is caused by pressure gradients. Air is accelerated in direction of the velocity if the pressure goes down. Thus the decrease of pressure is the cause of a higher velocity. [See also Weltner, Klaus; Ingelman-Sundberg, Martin, Misinterpretations of Bernoulli's Law]
Let's now focus on your two examples:
- The passing train causes an increase in the velocity of the air in proximity of the platform edge. This temporary effect, lowers the pressure experienced in front of a person standing there, i.e., it creates a difference of pressure between the back and the front of the person, pushing her forward.
To be more precise, the pressure of the air is $p_A$ before the approaching train passes. Since everything is at equilibrium, there is no gradient between the air in front and behind the edge of the platform. During the transit, the air between the person and the train experiences an increased velocity, and therefore a lower pressure, e.g. $p'=p_A-\Delta p$. This $\Delta p$ can be calculated using Bernoulli's equation. Its effect is that of a net force, pushing the person from the edge of the platform towards the tracks.
- The same effect causes an house roof to blow off when there is strong wind. Indeed, when the wind kicks up, the pressure on top of the roof (outside the house) drops, and the pressure gradient between inside (higher pressure, stationary air) and outside (low pressure, high speed) works as a net force pushing the roof upwards.
In both cases one could argue that the air could flow in the surroundings of the place where the phenomenon takes place, restoring equilibrium. But in fact, in both cases, the velocities are such that there isn't time for air to flow in and out of the house to cause the corresponding decrease in the interior pressure. That's why you get a pressure gradient across the roof.
The technical answer
First I need to explain entropy. Suppose you have any system which you can only see at a certain granularity: we say that you see its "macrostate" but this could be any set of "microstates" which all "look alike". Since particle-interactions tend to multiply and distribute our uncertainty about a system, we could imagine under certain circumstances (constant total energy, constant volume, constant number of particles) that we are selecting a microstate essentially at random: hence the "largest" macrostate is the one that the system naturally "wants" to be in due to uncertainty-multiplication. ("Largest" here means: it contains the most distinct microstates. We would say that it has the most "phase space volume" where phase space adjoins momentum-space to real-space.) This additive measure of how large a macrostate is, is known as "entropy."
If a system has constant numbers of particles and is transitioning between states of equivalent entropy (so that randomness doesn't drive the transition), then the pressure is the rate of change of the total internal energy of the system, with respect to the volume. This could happen for whatever reason; it is all lumped together into the term "pressure."
Since energy is usually globally conserved by Noether's theorem, this is equivalent to saying "pressure is the capacity for a closed system to do work when it changes volume." This is probably the hardest theoretical thing to understand about pressure in the long run. It has a dual status: we can speak of the pressure at a point, but also distributed throughout a system. The full reconciliation of this dual nature involves treating each small cube of the fluid as a "grand canonical ensemble" which is sharing its particles and temperature with the much larger "system", and we can nevertheless ask how its "free energy" would vary with its volume, if it were to expand and then its internal energy and particle numbers equilibrated with its surroundings. That's a little more intense than this section allows.
But yeah, each point has a capacity to do work (in the form of a change in local free energy per change in local volume) and added up appropriately this means that a closed system has a capacity to do work (in the form of a change in total internal energy per change in total volume). Anything which contributes to this capacity is called "pressure".
What effects are lumped together under pressure?
The ideal gas does not have any self-interaction terms, and the pressure is solely a measure of kinetic energy per unit volume. This kinetic energy density causes the particles to push against the walls all the time, so that is how the pressure can do work. As we've said, the capacity to do work, from any source, shows a pressure.
Now let's take an ideal gas and turn on some particle-particle interactions. Let's consider a repulsive one as the easiest: imagine that we just suddenly gave each particle one electron of negative charge, so that they were all repelling each other.
The first effect, which has no effect on pressure, is that we had to secretly pump in a lot of internal energy to do this: we put a total charge $Q = N~e^-$ into the system, which changed the voltage of the box to $V$, so there may have been something like $VQ/2$ energy dumped into the box simply in this off-hand "throw some charge on it!".
But the secondary effects are more interesting. In conductors, charge tends to pile up on the edge of the box: so the center of the box now has a much lower density, the outside has a much higher density, and so we roughly would expect that the added "push" of the system outwards manifests as a higher total pressure. Repulsive particle interactions increase pressure, attractive particle interactions reduce it. You can similarly imagine that the attractive interaction means that when you increase volume, you get a "bump" from kinetic energy but you have to "tear apart" the potential energy holding these guys together, if that helps you visualize why the force on the external world is weaker.
Finally, it's worth considering diatomic compounds like $O_2$. These things can be treated a lot like ideal gases, but they have an internal energy (rotational kinetic energy) which doesn't tend to contribute to the pressure. This is to encourage you to forget the fallacy "average internal energy per unit volume" or some such; it's a rate-of-change, not an average.
Example: van der Waals equation of state
Probably the most famous example of the effects of particle-interactions on pressure is the so-called van der Waals equation. This is a simple, early heuristic to capture the non-ideal effects of a changing volume and pressure on a real fluid. It turns out that it contains a liquid-gas phase transition at a certain temperature, so it is our first stop also when we want to introduce phase transitions to our students. Actual fluids have been fitted to the following equation for parameters $(a, b)$:$$
\left( p + a~\left(\frac {n}{V}\right)^2\right)~\big(V - b~n\big) = n~R~T.
$$See for example, Wikipedia's data page of these constants $a$ and $b.$ Now I can roughly explain that this is trying to equate a "total energy" (left) with a "thermal energy" (right). The term $b$ refers to a repulsive short-range potential which keeps the particles from occupying the same location: in real atoms this is because the electron clouds do not want to overlap; it is being modeled by pretending that the particles are secretly hard spheres and therefore the "available" volume is not the "total" volume $V$ but instead decreases proportional to $n$. The term $a$ refers to an attractive longer-range potential which makes the particles want to stick together; as I said above, attractive forces should reduce pressure in favor of some sort of internal energy density. In detail, we can see that we modify $p \mapsto p + \alpha \cdot (\text{# of handshakes}),$ if we envision the particles as "shaking hands" with each other: so for calculating the equation of state, we take the lowered-pressure and bolster it by the amount it was reduced. Since this attractive force drops with distance, the number-of-handshakes is not calculated like $n^2/2$ (total number of handshakes throughout the volume) but instead $n^2 / V^2,$ (handshakes with nearby neighbors, for some definition of "nearby").
Then the $p-V$ diagram for lower temperatures has a clear "dip" where you have to increase the pressure to compress it (fighting $b$) but also to expand it (fighting $a$). This manifests as a sudden amount of energy you'd get once you put enough pressure on it, as it all transitions from a gas to a liquid.
So what is pressure energy?
Summing this all together, pressure energy is the energy contained in each unit of the fluid due to the effects of thermal kinetic motions of the atoms lessened by the attractive forces of the fluid molecules on each other. Even if the fluid is viewed as incompressible from the point of view of the flow (i.e. the fluid flow is much less than the speed of sound in the fluid) it still answers the question of "if we changed the local volume, how would the local free energy change?". More importantly, it drives the motion of particles from one place in the fluid to another: if you pressurize this side by having it in contact with a fixed volume of air which you're pumping more air into with a bike pump, the added pressure gradient in the fluid causes the fluid to flow out of the reservoir faster, and into whatever else the system is connected to.
Best Answer
The terms are not the averaged energy per volume as you derive for your container, but the energy per volume for an infinitesimally small parcel of liquid at some point in the liquid (and the equation is valid along a stream line of the liquid).