Understanding Feynman's development of the ideal gas law: Vol I 39-2 of The Feynman Lectures on Physics In The Feynman Lectures on Physics, Volume I 39-2 The pressure
of a gas, the following
is presented:

If $v$ is the velocity of an atom, and $v_{x}$ is the $x$-component
  of $v$, then $mv_{x}$ is the $x$-component of momentum in;
  but we also have an equal component of momentum 0ut and so the total momentum delivered to the piston by the particle, in one collision, is $2mv_{x}$, because it is "reflected".
Now, we need the number of collisions made by the atoms in a second,
  or in a certain amount of time $dt;$ then we divide by $dt$. How
  many atoms are hitting? Let us suppose that there are $N$ atoms in
  the volume $V$, or $n=N/V$ in each unit volume. To find how many
  atoms hit the piston, we note that, given a certain amount of time
  $t$, if a particle has a certain velocity toward the piston it will
  hit during the time $t$, provided it is close enough. If it is too
  far away, it goes only part way toward the piston in the time $t$,
  but does not reach the piston. Therefore it is clear that only those
  molecules which are within a distance $v_{x}t$ from the piston are
  going to hit the piston in the time $t$. Thus the number of collisions
  in a time $t$ is equal to the number of atoms which are in the region
  within a distance $v_{x}t,$ and since the area of the piston is $A,$
  the volume occupied by the atoms which are going to hit the
  piston is $v_{x}tA$.  But the number of atoms that are going
  to hit the piston is that volume times the number of atoms per unit
  volume, $nv_{x}tA.$ Of course we do not want the number that hit
  in a time $t$, we want the number that hit per second, so we divide
  by the time $t$, to get $nv_{x}A$. (This time $t$ could be made
  very short; if we feel we want to be more elegant, we call it $dt,$
  then differentiate, but it is the same thing.)
So we find that the force is 
$$
F=nv_{x}A\times2mv_{x}.\,\,\,(39.3)
$$
See, the force is proportional to the area, if we keep the particle
  density fixed as we change the area! The pressure is then 
$$
P=2nmv_{x}^{2}.\,\,\,(39.4)
$$
Now we notice a little trouble with this analysis: First, all the
  molecules do not have the same velocity, and they do not move in the
  same direction. So, all the $v_{x}^{2}$'s are different! So what
  we must do, of course, is to take an average of the $v_{x}^{2}$'s,
  since each one makes its own contribution. What we want is the square
  of $v_{x}$, averaged over all the molecules: 
$$
P=nm\left\langle v_{x}^{2}\right\rangle .\,\,\,(39.5)
$$
Did we forget to include the factor 2? No; of all the atoms, only half are headed toward the piston. The other half are headed the other way, so the number of atoms per unit volume that are hitting the piston is only $n/2$.

While I accept the result, I do not understand his development. In
particular, what is meant by the "volume" $v_{x}tA$? That volume
is introduced with the tacit (and incorrect) assumption that all $v_{x}$'s are equal.
An assumption which is subsequently rejected. But the meaning of $v_{x}tA$
in terms of the refined understanding of $v_{x}$ being specific to
each atom is never made clear. 
Introducing the notation $\Delta V_{x}=v_{x}tA;$ I have found no
way to arrive at the advertised result $\left(39.5\right)$ using
half the average $x$-component of speed to establish the correct
value of $\Delta V_{x}$. For example:
$$
\frac{1}{2}n\left\langle \left|v_{x}\right|\right\rangle A\times2m\left\langle \left|v_{x}\right|\right\rangle =Anm\left\langle \left|v_{x}\right|\right\rangle ^{2}\ne Anm\left\langle v_{x}^{2}\right\rangle .
$$
The result $\left(39.5\right)$ can be established by an alternative
development which determines the number of collisions per unit time
by considering the number of times any specific particle will traverse
the $x$-dimension of a box of unit volume to the far side, and then
back in a time $t$. But I am interested to know if Feynman's approach
can be understood.
Under the assumption that the value $v_{x}$ is specific to each atom,
what volume, corresponding to the above, $\Delta V_{x}=v_{x}tA$, should be used
to determine the number of collisions per unit time of gas atoms with
the piston?
 A: Well, first thing I would like to say is that you are expecting too much rigor from a semi-handwaving demonstration such as this one.
Anyway, what you can argue is that you expect the distribution of $|v_x|$ to have a small (relative) variance, i.e.
$$\frac{\langle |v_x|^2 \rangle - \langle |v_x| \rangle^2}{\langle |v_x| \rangle^2} = \frac{\langle |v_x|^2 \rangle}{\langle |v_x| \rangle^2} -1 \ll 1 $$
Now of course in principle you don't know if this is reasonable. If you actually use the Maxwell-Boltzmann distribution you will discover that 
$$\frac{\langle |v_x|^2 \rangle}{\langle |v_x| \rangle^2} -1 = \left(\frac{kT}m\right)\left(\frac{\pi m}{2kT}\right)-1 = \frac \pi 2 -1\approx 0.57 $$
So things are not so good, but not so bad either (we got a number smaller than $1$, but not by much...).
In the end, I would say that you are right in thinking that there is a bit too much handwaving in this proof. Probably, as it often happens in such rough estimates, there is some fortuitous cancellation of errors: you overestimate the value of something, but underestimate the value of something else and in the end you get the correct result. This happens all the time in physics (look at the Drude model for conduction or at the Flory theory of polymers, for example).
Notice that the derivation that you can find for example on Wikipedia is slightly different from Feynman's derivation. The crucial difference is that Feynman estimates the number of collisions per second $f$ as
$$f=\frac{n V(\Delta t)}{\Delta t}$$
The problem is the use of the $V(\Delta t)$, which as you pointed out can only be reasonably estimated using average quantities:
$$V(\Delta t) = A \langle |v_x | \rangle \Delta t$$
In the Wikipedia derivation, a more stringent assumptions are made: the box is a cube of length $L$.
Instead of estimating $f$ using this "average volume" $V(\Delta t)$, it is simply stated that the frequency of the collisions is
$$f=\Delta t^{-1} = \frac{v_x}{2L}$$
where $\Delta t$ is the time it takes a particle to go form one side of the box to the other moving in the $x$ direction.
Then the force is calculated as
$$F= f \Delta p = \frac{v_x}{2L} \cdot 2mv_x = \frac{mv_x^2}{L}$$
Since the two $v_x$ that appears in the previous formula are referred to exactly the same molecule, there is no $\langle |v_x | \rangle$ coming into play and we can safely say, after averaging over all the molecules, that the average force will be 
$$\langle F \rangle = \frac{m \langle v_x^2 \rangle}{L}$$
Notice that in this case we don't even have to correct for the factor $2$ which comes out from Feynman's argument. So we get the "right" result without any ambiguity regarding $\langle |v_x|\rangle^2$ vs $\langle v_x^2 \rangle$. The price to pay, however, is that we had to sneak in an additional assumption.
