$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\cn}{cn}\DeclareMathOperator{\am}{am}$
I looked a little at the first video (William Schwalm's lecture). There he uses an angle $\theta$ which is such that the point $P = (x,y)$ on the ellipse $(x/a)^2 + y^2 = 1$ (with $a>1$, and hence eccentricity $k = \sqrt{a^2-1}/a$) satisfies
$$
(x,y) = (a \cn(u),\sn(u)) = (r \cos\theta,r \sin\theta)
.
$$
This angle $\theta$ is clearly not the same as the classically used angle $\phi = \am(u,k)$ which is such that $u = F(\phi,k)$ and
$$
(\cn(u),\sn(u)) = (\cos\phi,\sin\phi)
.
$$
In fact, from these formulas it follows that
$$
\tan\theta = \frac{\sn(u)}{a \cn(u)} = \frac{1}{a} \tan\phi
.
\tag{$*$}
$$
So you're misquoting his definition by putting the angle $\phi$ where he actually uses $\theta$.
Schwalm defines $u$ as a function of $\theta$ through the differential relation $du = r \, d\theta$.
Let us check that this does agree with the classical definition, where we have
$$
\frac{du}{d\phi} = \frac{d}{d\phi} F(\phi,k) = \frac{1}{\sqrt{1 - k^2 \sin^2 \phi}}
$$
from the fundamental theorem of calculus.
To begin with, using $a^2 = 1/(1-k^2)$ we have
$$
\begin{split}
r^2 &
= x^2 + y^2
= a^2 \cn^2(u) + \sn^2(u)
\\ &
= a^2 \cos^2 \phi + \sin^2 \phi
= \frac{\cos^2 \phi}{1-k^2} + \sin^2 \phi
= \frac{1 - k^2 \sin^2 \phi}{1-k^2}
.
\end{split}
\tag{$**$}
$$
Next, from the identity $(*)$ above, we obtain
$$
\begin{aligned}
\frac{d}{du} \tan\theta &= \frac{1}{a} \frac{d}{du} \tan\phi
\\ \iff \quad
(1 + \tan^2 \theta) \frac{d\theta}{du} &= \frac{1}{a} (1 + \tan^2 \phi) \frac{d\phi}{du}
\\ \iff \quad
(1 + a^{-2} \tan^2 \phi) \frac{d\theta}{du} &= \frac{1}{a \cos^2 \phi} \frac{d\phi}{du}
\\ \iff \quad
\frac{du}{d\theta}
= \frac{1}{d\theta/du}
&
= (1 + a^{-2} \tan^2 \phi) (a \cos^2 \phi) \frac{1}{d\phi/du}
\\ &
= a (\cos^2 \phi + a^{-2} \sin^2 \phi) \frac{du}{d\phi}
\\ &
= (1-k^2)^{-1/2} (\cos^2 \phi + (1-k^2) \sin^2 \phi) \frac{1}{\sqrt{1 - k^2 \sin^2 \phi}}
\\ &
= \frac{\sqrt{1 - k^2 \sin^2 \phi}}{\sqrt{1-k^2}}
= r
,
\end{aligned}
$$
where $(**)$ was used in the very last step,
so that indeed $du = r \, d\theta$, as claimed.
From $(*)$ we also get $a^2 \tan^2 \theta = \tan^2 \phi = \sin^2 \phi / (1 - \sin^2 \phi)$,
which after a bit of calculation tells us that
$1 - k^2 \sin^2 \phi = (1-k^2) / (1 - k^2 \cos^2 \theta)$,
so that
$$
r = \sqrt{\frac{1 - k^2 \sin^2 \phi}{1-k^2}} = \frac{1}{\sqrt{1 - k^2 \cos^2 \theta}}
.
$$
(This can also be seen, perhaps more easily, by inserting $(x,y)=(r \cos\theta, r\sin\theta)$ into the ellipse's equation $(x/a)^2+y^2=1$.)
Thus, the relationship between $u$ and $\theta$ can be written as
$$
u = \int_0^\theta \frac{d\alpha}{\sqrt{1 - k^2 \cos^2 \alpha}}
,
$$
to be contrasted with the relationship between $u$ and $\phi$,
$$
u = F(\phi,k) = \int_0^\phi \frac{d\beta}{\sqrt{1 - k^2 \sin^2 \beta}}
.
$$
In the current version (Feb 2023) of the Wikipedia article for
Jacobi elliptic functions,
there is a section
Definition as trigonometry: the Jacobi ellipse,
which contains a construction involving the ellipse $x^2 + (y/b)^2 = 1$ with $b>1$.
Despite the superficial similarity, and several references to Schwalm on the
the article's talk page,
this construction is not the same as the one in the video,
since they write an integral producing a point $(x,y) = (r \cos\phi, r \sin\phi)$
on the ellipse, where $\phi$ (written $\varphi$ on Wikipedia) is the classical angle (not $\theta$),
and then they project that point radially from the ellipse to the unit circle in order to get the
point $(\cn u, \sn u) = (\cos\phi,\sin\phi)$.
That is, they are not defining $\cn$ or $\sn$ directly as the $x$ or $y$ coordinate of some point on their ellipse, as Schwalm does with $\sn$ on his ellipse (he is projecting his point $(x,y)$ horizontally from the ellipse to the unit circle in order to get the
point $(\cn u, \sn u) = (\cos\phi,\sin\phi)$).
