I have already read your post and the answers to the questions you linked before. While I completely agree that the explanations given do not really answer the questions, I am not quite sure whether I will be able to give an answer that does. Nonetheless I will give it a try.
Fluids: Liquids and gases
Indeed liquids and gases behave practically identical on a macrosopic level, they are both continua characterised by viscous damping behaviour and can be approximated (apart from some exotics such as Bingham fluids) as Newtonian fluids. As a consequence the governing macroscopic laws, the conservation of macroscopic quantities (e.g. mass, momentum and energy), such as the Navier-Stokes equations take the identical form
$$\frac{\partial \rho^*}{\partial t^*} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho^* u_j^* )}{\partial x_j^* }=0,$$
$$\rho^* \frac{\partial u_i^*}{\partial t^*} + \rho^* \sum\limits_{j \in \mathcal{D}} u_j^* \frac{\partial u_i^*}{\partial x_j^*} = - \frac{\partial p^*}{ \partial x_i^* } + \frac{1}{Re} \sum\limits_{j \in \mathcal{D}} \frac{\partial \tau_{ij}^*}{\partial x_j^* } + \frac{1}{Fr^2} g_i^*,$$
$$\rho^* \frac{\partial T^*}{\partial t^*} + \rho^* \sum\limits_{j \in \mathcal{D}} u_j^* \frac{\partial T^*}{\partial x_j^*} = Ec \left( \frac{\partial p^*}{\partial t^*} + \sum\limits_{j \in \mathcal{D}} u_j^* \frac{\partial p^*}{\partial x_j^*} \right) + \frac{1}{Pr Re} \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j^*} \left( \frac{\partial T^*}{\partial x_j^*} \right) + \frac{Ec}{Re} \sum\limits_{i, j \in \mathcal{D}} \tau _{ij}^* \frac{\partial u_i^*}{\partial x_j^*} ,$$
just the characterstic dimensionless numbers take different orders of magnitude (certain effects dominate over others). This is somewhat puzzling given that on a microscopic level particles of liquids and gases are assumed to interact quite differently: Gases can be assumed to be small separated particles or molecules while liquids are more dense and may involving large highly asymmetric molecules and interaction mechanisms can be significantly more complex with repulsive and attractive forces. This fact is though taken into account on a macroscopic level by the completely different equations of state of liquids and gases (required to close the equation system) and the different order of magnitude of the dimensional numbers.
Pressure as a macroscopic variable
As you can see it does not really seem to matter what you consider, a dense liquid or a comparably dilute gas with simplified interactions: In the limit of small Knudsen numbers both behave identically. Similarly simplified gas models such as automata that describe collision rules on a microscopic level can yield the ordered behaviour of a gas. Nonetheless one is unlikely to give a general explanation for properties such as pressure on a microscopic level that is valid for all kinds fluids: The similar properties on macroscopic level emerge from different microscopic mechanisms that mainly have one thing in common - damping.
On a macroscopic level pressure is nothing more than a force per area. It has to be in balance with the forces around it. Pressure can take several forms that are all a consequence of a certain force per area: The momentum flux stemming from macroscopic motion is termed dynamic pressure (that's the contribution you mainly feel when you stick the hand out of your car while driving on the highway),
$$ p_d = \frac{\rho u_i u_i}{2} $$
while the isotropic pressure (in all directions the same) that determines the properties of a fluid (e.g. in the equation of state) is also termed static pressure $p$. Clearly if there is a certain liquid column above a certain point this exerts a force as well, characterised by the hydrostatic pressure $p_h = \rho g h$, that also contributes to the static pressure. The combination of both static and dynamic pressure is often referred to as total pressure or more correctly stagnation pressure because that's the pressure you feel in a stagnation point of the flow (at least if you slow the velocity down to zero isentroptically).
$$ p_s = p + p_d.$$
The hydrostatic pressure has a direct influence on the equation of state. As you can clearly see incompressibility has to be compatible with the equation of state!
Kinetic theory of gases
At the end of the 19th century Maxwell and Boltzmann almost single-handedly established the field of "kinetic theory of gases". Already before it was known that the world is composed by atoms both tried to describe a gases as a collection of interacting particles. Already very simple analytical models such as the 1/6 model are able to estimate transport quantities in (dilute) gases and probably motivated by these findings in particular Boltzmann tried to describe a dilute gas as a multi-body system interacting in collisions, using newly developed tools such as statistical mechanics.
The view of a continuum is based on the assumption that you can find macroscopic properties such as density or pressure that require a sufficient amount of particles such that those limiting values exist and are sufficiently smooth (in space and time). If you don't (indicated by high Knudsen numbers $Kn := \frac{\lambda}{L}$ where $\lambda$ is the mean free path) you are screwed with your macroscopic view, it will simply fail.
Assuming that the world is composed of individual particles with their respective velocities one might still find a certain probability that a particle with a certain speed exists in a certain phase space volume: Around a certain point in space $\vec x$ you might find a particle within a certain velocity interval $\vec \xi$ with a certain probability $f$
$$f = \frac{d N}{ d \vec x \, d \vec \xi}.$$
The macroscopic quantities, density, momentum and total energy, emerge as expected values
$$\rho = m_P \int\limits_{\vec \xi} f d \vec \xi, \hspace{1cm} \rho u_i = m_P \int\limits_{\vec \xi} f \xi_i d \vec \xi, \hspace{1cm} \rho \left( e_i + \frac{u_i u_i}{2} \right) = m_P \int\limits_{\vec \xi} f \xi_i \xi_i d \vec \xi.$$
We could now try to find an evolution equation for this probability. Likely motivated by Hamiltonian mechanics and the Louiville equation Boltzmann tried to combine the evolution with a certain term describing the redistribution due to binary collisions based on elastic collisions that can be described my Newtonian mechanics, furthermore assuming molecular chaos (the two interactions are assumed to be uncorrelated before collisions), the Stosszahl ansatz, resulting in the Boltzmann equation
$$\underbrace{ \frac{\partial f}{\partial t} + \vec \xi \boldsymbol{\cdot} \vec \nabla f + \frac{\vec F}{m} \boldsymbol{\cdot} \vec \nabla_{\vec \xi} f }_\text{Propagation} = \underbrace{ = \int\limits_{ \vec \xi_1 } \int\limits_{ A_c } |\vec g| (f_1' f' - f f_1 ) d A_c d \vec \xi_1 }_\text{Collision} .$$
One can now ask him-/herself: Is there a certain attractor, a certain distribution that a system evolves to? And surprisingly already by considering symmetries and conservation of moments you can find the Maxwell-Boltzmann equilibrium distribution $f^{(eq)}$ and prove with the Stosszahl ansatz that a system evolves towards it over time and find a model entropy.
We could now try to express what this distribution does in terms of the macroscopic variables, how the system evolves on larger length and time scales. One way of doing so is the perturbation theory by Chapman-Enskog (something that traditionally stems from celestial mechanics and looks at an orbit that is slightly perturbed, so you decompose the solution $f$ into different contributions $f^{(n)}$ with different orders of magnitude $\epsilon$).
$$ f = \sum\limits_{n=0}^{\infty} \epsilon^n f ^{(n)}$$
Surprisingly in the limits of dense fluids leads to the Euler equations considering only the first term $f^{(0)} = f^{(eq)}$ and to the full Navier-Stokes equations if you consider the following two terms as well. You can find certain terms for transport coefficients that connect the "microscopic" distribution to properties of the fluid on a macroscopic level like viscosity.
Note: This simplified model is motivated by a dilute model gas and its limiting value might be argued to be a dense gas not a liquid. Furthermore the interactions are more simple than in real gases, e.g. vibrational degrees of freedom are not considered.
Pressure on a microscopic level in the kinetic theory of gases
With a model such as the kinetic theory of gases that is based on collisions you are able to say more about what pressure actually looks like on a microscopic level. You can find
$$\lambda = \frac{m_P}{\sqrt{2} \pi d^2 \rho}$$
to correspond to the mean free path in a dilute gas. You can instantly see that the mean free path is inversely proportional to the density: A higher density means a lower mean free path and thus a shorter distance between collisions of particles. The density is connected to the pressure via the equation of state of an ideal gas in this case
$$p V = k_B T = n R T = N k_B T$$
As you can see using $n := \frac{m}{M}$ and $\rho := \frac{m}{V}$
$$p \, v = \frac{p}{\rho} = R_m T$$
a higher pressure also corresponds to a higher density $\rho$ and thus also to a lower mean free path. This means again that you do not need a temperature gradient in an ideal gas to fulfill and explain the hydrostatic pressure. The fluid column above in an ideal gas presses the molecules closer together reducing the density and thus leading to a higher static pressure: Particles are not faster (as you would expect for higher temperature) but there are simply more particles hitting an area element exchanging a larger amount of momentum.
Incompressible fluids and hydrostatic pressure
As in the post mentioned above the term incompressible always is accompanied by misunderstanding and confusion. I have written a post on incompressible fluids and incompressible flows some time ago if you are interested. Incompressibility is an artificial concept which needs a physical motivation and clearly has to be compatible with the equation of state and the flows itself!
Arguing about pressure in a microscopic context considering the findings of the kinetic theory of gases, incompressible fluids and hydrostatic pressure is a contradiction by itself. An incompressible fluid requires the density $\rho$ to be constant! As you can instantly see this would require a temperature gradient in order to fulfill the equation of state of an ideal gas if the pressure varies greatly! A gas can be assumed incompressible in some limits but not if the weight of the fluid above compresses the gas on the bottom significantly so the density is not approximately homogeneous (You might apply incompressibility to a vehicle moving approximately on a iso-density surface but not in the direction perpendicular to it as the pressure will vary greatly!)
In a liquid this is different. The equation of state for water is generally given by the Tait equation
$$p - p_0 = C \left[ \left( \frac{\rho}{\rho_0} \right)^m - 1 \right] $$
where the exponent $m$ is estimated to be around $7$. This means that in a liquid assuming incompressibility holds even for very high liquid columns, as a small change in density leads to a huge change in pressure.
Microscopic view: Pressure in liquids
A potential way of thinking of a liquid is thinking of it as a collection of particles that is so dense that you barely can compress them any more (incompressible fluid). As a result external forces will not lead to a compression of the liquid itself but instead will only increase the force and thus the pressure between the densely-packed particles. While already the kinetic theory of gases is a huge simplification (There exist some really complicated concepts that take far-field interactions into account as well!) of the actual physics, this model is even more so, as longer molecules allow for more complicated interactions (think of hydrogen bonds).
