The more formal derivation of the van der Waals equation of state utilises the partition function. If we have an interaction $U(r_{ij})$ between particles $i$ and $j$, then we can expand in the Mayer function,
$$f_{ij}= e^{-\beta U(r_{ij})} -1$$
the partition function of the system, which for $N$ indistinguishable particles is given by,
$$\mathcal Z = \frac{1}{N! \lambda^{3N}} \int \prod_i d^3 r_i \left( 1 + \sum_{j>k}f_{jk} + \sum_{j>k,l>m} f_{jk}f_{lm} + \dots\right)$$
where $\lambda$ is a convenient constant, the de Broglie thermal wavelength and this expansion is simply obtained by the Taylor series of the exponential. The first term $\int \prod_i d^3 r_i$ simply gives $V^N$, and the first correction is simply the same sum each time, contributing,
$$V^{N-1}\int d^3 r \, f(r).$$
The free energy can be derived from the partition function, which allows us to approximate the pressure of the system as,
$$p = \frac{Nk_B T}{V} \left( 1-\frac{N}{2V} \int d^3r \, f(r) + \dots\right).$$
If we use the van der Waals interaction,
$$U(r) = \left\{\begin{matrix}
\infty & r < r_0\\
-U_0 \left( \frac{r_0}{r}\right)^6 & r \geq r_0
\end{matrix}\right.$$
and evaluate the integral, we find,
$$\frac{pV}{Nk_B T} = 1 - \frac{N}{V} \left( \frac{a}{k_B T}-b\right)$$
where $a = \frac23 \pi r_0^3 U_0$ and $b = \frac23 \pi r_0^3$ which is directly related to the excluded volume $\Omega = 2b$.
(a) $$\sum \text{translational KEs of molecules} =N \times \tfrac12 m c_{rms}^2=\tfrac12 N m c_{rms}^2$$
But
$$pV=\tfrac13 N m c_{rms}^2$$
So the correct relationship for an ideal gas is
$$\text{Total translational KE} =\tfrac32pV$$
(b) Dalton's law states that the pressure exerted by a mixture of gases (in a container) is the sum of the pressures that would be exerted by each gas if it were the only gas in the container. So the law doesn't say anything about what happens when you mix gases at different temperatures. The gases that the law deals with are already mixed and therefore at the same temperature!
(c) When you talk of your textbook proving Dalton's law, you mean, I think, deriving it from the kinetic theory of ideal gases. You can do the derivation without considering the sum of molecules' kinetic energies. You simply extend the assumption of additivity of momentum changes due to molecules' impacts with the wall to include molecules of different species but with the same $m c_{rms}^2$.
(d) At the end of your question you assume that when you mix two gases at different temperatures the temperature rise of one gas is equal to the temperature fall of the other. This will not, in general, be true. Just think of what would happen if you mix 10 mole of a hot gas with 1 mole of a colder gas. It is, of course, the case that both gases will end up at the same temperature.
(e) How should one find an expression for the final pressure of equal volumes, $V$, of two gases at different temperatures and pressures when mixed together and confined to a volume, $V$?
You know that the number of moles is conserved. Hence, using the ideal gas equation, you can find a relationship involving just pressures and temperatures.
But the confinement won't take place spontaneously, as you suggest: "Consider another container with the same volume and slowly mix these both gases from those containers into this new one." You'll have to do work to squash 2 volumes $V$ of gas into volume $V$. This stops internal energy being conserved even if the mixing is adiabatic, so a nice easy equation is denied you.
But internal energy would be conserved if the final volume of the mixture were allowed to be $2V$, provided that the mixing were adiabatic. If you further assume that the molar heat capacity of the two gases is the same, you get a very simple result for the final pressure in terms of the initial pressures.
Best Answer
Firstly, $P$, $V$ and $T$ in the van der Waals equation of state have exactly the same meanings as they have elsewhere in thermodynamics: they represent the pressure, volume, and temperature of a sample of the material being studied.
Secondly, this equation of state is one of many approximate equations of state which attempt to model the values measured in experiment, especially (in this case) non-ideal gases. There are many other, more accurate, equations of state which are empirical fits to the measured data. The van der Waals equation is a useful one to study because it is simple, and one can understand in fairly simple terms where it comes from. It has some basis in statistical mechanics, which also helps explain its form.
But remember, it is just an approximation. For example, although it predicts the existence of a critical point, it continues in an unphysical way below the critical temperature into the two-phase region where liquid and gas are in equilibrium with each other.
The equations of state page actually gives a handwaving explanation of the two modified terms. I'll paraphrase here.
Indeed $(V-nb)$ attempts to account for the excluded volume of the atoms. If we ignore all the attractions, we are essentially modelling the atoms as hard spheres, and writing the equation of state as $$ \frac{PV}{nRT} = \left( \frac{V}{V-nb}\right) = \left( \frac{1}{1-b\rho}\right) $$ where $\rho=n/V$. This means that, for a given density $\rho$ and temperature $T$, the pressure is higher than we might have expected if we had assumed that the gas were ideal. Also, it recognizes that there is a minimum possible value of $V/n$ or a maximum density $\rho$, because at some point the atoms will overlap. (We don't take the precise value of the maximum density too literally; in practice, the system will crystallize to a solid, which the equation does not describe). Nowadays, we know a lot more about the hard sphere fluid, and have more accurate equations of state such as Carnahan-Starling for this system.
The effects of attractions are then treated as a weak perturbation, superimposed on the above equation. The gas surrounding a typical atom is treated as a uniform attractive background: a kind of mean field theory. This reduces the pressure, compared with what we might have thought on the basis of the ideal gas equation. The equations of state page gives a way to think of this:
In fact, this can be made more rigorous: it can be shown using that, for thermodynamic consistency, if we add a uniform attractive background to a system, the effect on the equation of state must take the form $$ P = P_0 - a\rho^2 $$ where $P_0$ is the pressure of the system without the attractions (for example, $P_0=1/(1-b\rho)$ as above, or a more accurate equation such as Carnahan-Starling) and $a$ is a positive constant. This is the form appearing in van der Waals' equation.
[EDIT following OP comment]
If the van der Waals equation describes the real gas accurately, then $P$ is the pressure of the real gas (not an ideal gas) and likewise $V$, $T$, $n$ are properties of the real gas. If you want to compare the properties of a real gas and an ideal gas, then it is necessary to specify what is kept the same, and what is to be compared. I believe that it is clearest to keep $n$, $V$ and $T$ the same, and compare pressures. Then, rearranging the equation of state \begin{align*} P &= \frac{nRT}{V-nb} - a\frac{n^2}{V^2} \\ \frac{PV}{nRT} = \frac{P}{P_{\text{id}}} &= \frac{V}{V-nb} - \frac{a}{RT}\frac{n}{V} = \frac{1}{1-\rho b} - \frac{a}{RT}\rho \end{align*} where I have introduced the density in moles per unit volume, $\rho=n/V$, and $P_{\text{id}}=nRT/V$ which is the ideal gas pressure at the same density and temperature. At low density, $\rho\rightarrow 0$, $P\rightarrow P_{\text{id}}$, as we would expect. The equation makes clear that the $b$ term (excluded volume due to short range repulsive interactions) tends to increase $P/P_{\text{id}}$ and the $a$ term (attractive long range interactions) tends to decrease $P/P_{\text{id}}$. Depending on the temperature and the density, it is possible for $P$ to be higher, or lower, than $P_{\text{id}}$.