To be consistent with the question, I'll use your notation rather than the notation in the video (which used $N$ where you use $q$ and $M$ where you used $n$).
The key to the limit of $\frac{q-1}{n}$ is that it is something that is true
when $n,q$ "are both equally large".
The idea is that $n$ depends on $q$ in some way such that the limit
$$ \lim_{n\to\infty} \frac qn = 1. $$
That's what it means for two numbers to be "equally large" in an asymptotic sense.
When this is true, it is also true that $\lim_{n\to\infty} \frac {q-1}n = 1.$
Personally, I think this is a very silly way to go about it. Let's just cut to the chase: if we set $n = q-1$ every time, it provides the desired simplification of the formula.
It also gives us a Riemann sum of the integral with $\Delta x=1.$
That is what the simplified formula calculates.
This particular choice of $n$ as a function of $q$ also satisfies the condition that $q$ and $n$ are asymptotically "equally large," for whatever that's worth.
Now we come to the choice of $x_k$. The graph in the video seems to be indicating that we choose each $x_k$ somewhere in the middle of its interval in the Riemann sum. But the actual formula for $x_k$ indicates that we are choosing the rightmost point in the interval.
That is, the given formula, $1 + k\left(\frac {q-1}n\right),$
simplifies to $1 + k$ if $n = q-1$,
which means that $x_1 = 2$ (for the rectangle between $x=1$ and $x=2$),
$x_2 = 3$ (for the rectangle between $x=2$ and $x=3$),
and $x_n = q$ (for the rectangle between $x=q-1$ and $x=q$).
The Riemann sum then tells us that
$$ \int_1^q \ln (x)\,\mathrm dx \approx \sum_{k=1}^n \ln(1 + k). $$
Next we have the remarkable claim that since $n$ and $q$ are "equally large,"
we can just change $n$ to $q$ for the upper index of the sum.
In the case where $n = q-1$, that's saying the approximation is just as good if you add $\ln(1+n)$ to the right-hand side.
But even if you say $n$ is not exactly $q-1$ but is just "equally large,"
it still is true that if $1$ is any kind of good approximation for $\frac{q-1}{n}$
(as required to make the video's argument work)
then replacing $n$ with $q$ in the sum means that we are increasing the sum by approximately $\ln(1+n).$
The video also messes up the re-indexing, because in order to change $\ln(1+k)$ to $\ln(k)$ while keeping the indexing the same, you have to pretend that it doesn't matter whether the term $\ln(1+q)$ is included in the sum or not.
A better way to deal with the sum is to write
$$ \int_1^q \ln (x)\,\mathrm dx \approx \sum_{k=1}^{q-1} \ln(1 + k), $$
that is, since we must insist that $\frac{q-1}{n}$ is approximately $1$ to make this all work, let's just say it is exactly $1,$ that is, $n = q-1$ (as I proposed earlier), and therefore it is perfectly OK to replace something (in this case, $n$)
with something exactly equal (in this case, $q-1$).
Now for the reindexing. For each value of $k,$ let $j = k + 1.$
Then as $k$ runs over the integers from $1$ to $q-1,$
$j$ runs over the integers from $2$ to $q$; and of course $\ln(1+k) = \ln(j).$
So we find that
$$ \sum_{k=1}^{q-1} \ln(1 + k) = \sum_{j=2}^{q} \ln(j). $$
But what if we want $j$ to start at $1$ instead of $2$? That just means we have
an extra term in the sum:
$$ \ln(1) + \sum_{j=2}^{q} \ln(j) = \sum_{j=1}^{q} \ln(j), $$
and since $\ln(1)=0,$ the left-hand side is just equal to $\sum_{j=2}^{q} \ln(j).$
Putting it all together,
$$ \sum_{k=1}^{q-1} \ln(1 + k) = \sum_{j=1}^{q} \ln(j). $$
Therefore
$$ \int_1^q \ln (x)\,\mathrm dx \approx \sum_{j=1}^{q} \ln(j). $$
Now we can simply rename the index variable from $j$ to $k$ in the sum,
and you have the final approximation shown in the video,
except that we got there without introducing one silly error and then introducing another error that just happens to cancel the first one.
Evaluating the integral exactly, we get
$$ \int_1^q \ln (x)\,\mathrm dx = q\ln(q) - q + 1 $$
(not $q\ln(q) - q$ as claimed in the video).
So if we accept then that
$$ \int_1^q \ln (x)\,\mathrm dx \approx \sum_{k=1}^{q} \ln(k) = \ln(q!), $$
this says that
$$ q\ln(q) - q + 1 \approx \ln(q!). $$
But Stirling's approximation is usually stated as an approximate formula for $q!,$
not $\ln(q!).$
Taking the exponential function of both sides, we get
$$ q! \approx \frac{q^q e}{e^q} = q^q e^{-q+1}. \tag1$$
If we use the final approximation in the video, we get
$$ q! \approx \frac{q^q}{e^q} = q^q e^{-q}. \tag2$$
Note that formula $(1)$ is $e$ times as large as formula $(2)$.
Try comparing the values produced by these two formulas with the actual value of $q!$ for a few values of $q$ and see how good an approximation you think they give.
The formula that is usually given as Stirling's approximation is
$$ q! \approx \left(\sqrt{2\pi q}\right) q^q e^{-q}. \tag3$$
Note that for any $q \geq 2,$ this formula is larger than either approximation $(1)$ or approximation $(2).$
In fact, it's about $0.922 \sqrt n$ times as large as approximation $(1)$,
so approximation $(1)$ is actually not very good
(it's off by a factor of $9$ for $n=100$) and approximation $(2)$ is even worse.
As it happens, $\ln x$ is an increasing function of $x,$ so the rectangles chosen for
the Riemann sum in the video
(where the upper right corner of the rectangle is on the curve)
are all larger than the corresponding areas under the curve.
The right-hand side is therefore an overestimate, never an underestimate
of the integral.
Again, since the function is increasing, the error is bounded by $\Delta x$ times the total increase of the function, that is,
the amount by which $\ln(q!)$ overestimates the integral is between $0$ and $\ln(q).$
The factor $\sqrt{2\pi n}$ comes from an approximation of that error.
