The fact that $xdist = a/2$ is a condition not an assumption. That must be the case for some angle of the diffraction, and when that is the case you can 'pair' up a portion of the beam with another portion of the beam and you will get complete diffraction. When $xdist$ does not equal $a/2$, then you cannot do this 'pairing up' and some portion of the beam will not destructively interfere. This how is the variation in intensity is explained.
Think about it this way: The a/2 condition essentially says that if you were to divide the beam up into two halves with an equally arbitrary resolution of beams in each half, then the first beam of the top half (at 0 measuring from the top of the slit) would destructively interfere with the first beam of the second half (at a/2 measuring from the top of the slit). Every beam in the top half would find a beam corresponding in the bottom half that would lead to destructively interference.
For a/4 we have the same case: We can split the beam up into 4 quadrants each containing an arbitrary number of beams. Beam 1 from quadrant 1 would destructively interfere with beam 1 from quadrant 3, etc. quadrant 1 and 3 destructively interfere. Same thing with quadrant 2 and 4, hence we also have a minimum here.
For a/5 it does not work: We split the beam up into 5 sections, with 4 of them we can make the same argumentation as before. That leaves us with 1 entire section that does not destructively interfere, no minimum.
What is the difference between a/2 and a/4: Think about it geometrically first. If we use the diagram from the video you posted, the a/4 condition would mean that we have the distance of $xdist=\lambda/2$ higher up and therefore the angle of the resulting beam would be angled higher. Turns out this is the condition for the second minimum. In fact, you will notice that this works with all multiples of 2, so let's make the algebraic:
$xdist=a*sin(\theta)=m*\lambda$ (where m is a non-zero positive integer)
This is the condition for all minima in the intensity pattern, i.e. places with complete destructive interference.
Since you now understand the geometrical shift of the minimum along with a, you can make sense of this condition optically: If a phase shift of $\lambda/2$ results in a destructive interference, then shifting that phase by a whole $\lambda$ should again result in a destructive interference.