Why is the slope of pressure and volume almost zero below critical point for liquefaction of gas?

The area under the curve is representing the region where the material is in both phases, i.e. it is in equilibrium between liquid and gaseous states, and is therefore extremely susceptible to small variations of thermodynamical properties.

Now, assume you are in a liquid state (coming from the "left" in your plot, e.g. from point D), and you enter the region of "instability" (following the line from point D you will enter the region at point C), where you have a liquid-to-gas phase transition. At a constant temperature, a small decrease in pressure is triggering the phase transition, and suddenly your material changes from liquid to gas. Since a gas is compressible, but a liquid is not, your material will experience a sudden increase in volume (ending up in point B). This is the reason why, when you move along the isotherms, the slope is almost horizontal: a phase transition from liquid to gas is accompanied by a sudden increase in volume, since the two phases have dramatic difference in compressibility.

On the molecular level, a very simple explanation is the following. A certain amount of pressure (at a certain constant temperature) is needed to keep the molecules in a liquid state. This external pressure is "fighting against" the kinetic energy of the molecules, keeping them "in place" (i.e. in a liquid state). If you decrease the pressure, the containing force keeping the kinetic energy of the molecules at bay is decreasing, and it will eventually happen that the kinetic energy of the molecules is strong enough to repel one another. When this happens the material is transitioning to a gaseous state, therefore undergoing a sudden expansion since its molecules are free to move and repel one another upon collision.

I hope this answer is not too reductive, but is helping to give the idea of the main mechanisms underneath the physics involved.

Maybe you heard the phrase that at the critical point there is no way to distinguish between the gas and the liquid phase. The question is: what changes?

Phases at different densities

This is because the key difference between the two is the difference between liquid density $rho_l$ and gaseous density $rho_g$. You can understand that on a molecular level. If the gas temperature is high enough, the molecules will move about freely in space, they have so much kinetic energy that the potential energy that they could gain from staying close to another (within an attractive region of the interaction potential) isn't really relevant to the motion. If you reduce kinetic energy (that is temperature), the wells of attraction in the interaction potential become more relevant and the molecules will want to stay close enough to one another so that they can energetically profit from the attractive interaction. Thus, we have a high temperature phase where molecules stay further away from each other - at lower densities $rho_g$ - and a low temperature phase where they move closer together to profit from attractive interaction - the high density phase with $rho_l$.

This is the process when one changes temeprature. Along an isotherm, a similar process can force the molecules apart or closer together as one alters the density. If the temperature is low enough, this will actually lead to two different phases, liquid and gas. At high densities, the system will try to profit from the attractive part of the interaction potentials and particles will want to stay close (the liquid phase), and at low densities, the largest contribution comes from particles evenly distributing over the volume to minimize entropy, which gives rise to the gas phase.

Phase separation

However, there is an intermediate region where phase separation occurs. Those are the horizontal lines in the coexistence region. In that region, the system features a liquid part that profits from attractive interactions, but this part alone cannot fill the whole container due to its higher density $rho_l > v^{-1}$. So the matter will separate into a liquid phase (if there is gravity, the liquid phase gathers at the bottom of the container, e.g. a kitchen pot) and a gaseous phase with $rho_g < v^{-1}$ that fills the rest of the container. This has something to do with the fact that the relevant quantity is the free energy $F = U -TS$ rather than the internal energy $U$ ($T$ is the temperature, $S$ the entropy), and a purely liquid system with areas of empty space wouldn't go well with the idea of maximum entropy. Notice that the densities of the two phases remain constant throughout the coexistence region. The overall change in density $v$ is due to more and more particles aggregating in one of the two phases. You can imagine that at high overall densities, the denser phase, the liquid, will be more prominent, and the opposite holds for high densities (note that density is inversely proportional to $v$). This whole process is known as liquid-gas coexistence.

Passing through the transition

Now to tackle your question, which is why there is no noticeable pressure change along an isotherm in the liquid-gas coexistence region. The key factor is that there is one very characteristic pressure: the pressure at the liquid-gas interface. Those two must be at the same pressure. Because if the liquid phaes were at a higher pressure than the gas phas, liquid particles would get "pressed out" of the liquid phase into the gaseous phase until the pressure equilibrates again. But at the same time, within each of the phases the temperature is constant (due to the isotherm we are moving on) and the density is, too, as we said above. So the pressure, for which we have a constitutive relation $P_{l/g} = P_{l/g}(T,rho_{l/g})$ cannot change, either, and must be equal to the density at the interface.

Therefore, if you wanted to change the pressure, you would have to go through the entire coexistence region, by evaporating all the liquid or liquefying all the gas. This process costs a lot of energy. When you increase temperature, however, the liquid and gas densities will both decrease, and the two regions "grow closer together", which you can see in your diagram by noticing that the coexistence region slims down as you go to higher temperatures. There is a temperature at which $rho_g = rho_l$. This is the critical temperature $T_C$. Above it, the attractive part of the interaction will not lead to large-scale condensed phases but only give rise to small-scale density fluctuations. Therefore, there are nor longer two macroscopic phases at different densities but just the one, and there is no longer phase coexistence. This is known as the critical transition that happens at the critical point.