Context
The wave equation takes the following form
$$\frac{\partial^2 f}{\partial t^2} = c^2\Delta f$$
where $c>0$.
This partial differential equation is usually for solutions on a given domain $\Omega$ and with given boundary and initial conditions.
If your drum membrane at rest as being a flat surface, then you can model it as $\Omega\in\mathbb R^2$. In that case, a solution $f(x,y,t)$ of the PDE represents the vertical displacement of point $(x,y)$ of the membrane at time $t$.
Why the solution of the 2D wave equation involves a double summation
Remember that $f(x,y,t)$ is a function of two coordinates $x$ and $y$, because the membrane itself is two dimensional, and one of time $t$. So there is no reason to expect a single Fourier summation as the function itself seemingly has 3 degrees of freedom. If anything you should expect 3 sums. However, because the time behavior is related to the space behavior by the wave equation, you can show that 2 sums are enough. But the bottom line is, it's more than just one sum.
To convince yourself, imaging the simplest case of the membrane being a flat rectangle at rest. The solutions to the wave equation can be written as $f(x,y,t)=u(x)v(y)w(t)$ as the equation is separable. And using Fourier series, the solution you get is the product of two Fourier series, which is a double sum over the Fourier modes in the $x$ and $y$ dimensions. The time oscillations are related to those same modes and the velocity $c$ (that explains why the 3rd sum is not needed).
Why the synthesis of sound via a drum or a speaker involves a single summation
That's because you're talking about the excitation function, not the solution of the wave equation. And you excite the drum in only one location (e.g. at its center $(0,0)$). Thus the excitation is a function of only one variable (time). Consequently, if you represent it via Fourier series or the Fourier transform, you'll only need one sum.
More details here, feel free to skip: To produce a sound, you need to excite the drum or speaker membrane. That means you need to provide it with energy. For a drum, this is done by hitting it with a drumstick. Now, think about what a simple quick hit of the drum looks like mathematically. It's short burst of energy passed on to the drum at a given specific location (where the drumstick hits the drum, typically the center). In this case, we usually model this by saying that the initial value of the displacement $f(x,y,0)$ is the Dirac delta. But this is for a very short burst of energy at $t=0$.
Next, imagine you want to excite the drum by hitting it at its center once every second. Then you can model it as a train to Dirac deltas.
Finally, instead of a simple pattern of hitting the drum once per second, you want to to continuously excite the drum, at least do it for some period of time. But the way you want to do it is by passing various amounts of energy, through its center, at different times. Then, this is like replacing the Dirac delta above, or the Dirac delta train, with a function $\varphi(t)$. This is exactly what happens for a speaker (= drum + a mechanism to continuously excite the membrane). Hopefully, by now, you see that the square wave function they mention in the tutorial is, very likely (I haven't watched the Youtube video, too long), the excitation signal. And by definition, this is a one-variable function (of time $t$), and can be represented by one sum only.
Conclusion
The various sums you are considering refer to 2 different objects:
- The waves on the drum membrane = function of 2 space variables and one of time, and needs double sums. You usually cannot reduce it to a single sum.
- The excitation function = function of 1 variable (time) and needs only one sum.
Consequently, there is no reason to expect that the two are equal.
Best Answer
You can represent a shifted square wave with a shifted Fourier series.
For example the shifted square wave with peak-to-peak amplitude $2$ and period $2 \pi$ defined as follows
$$f(t)=\left\{ \begin{array}{cc} -1 & -\frac{\pi }{4}\leq t<\frac{3 \pi }{4} \\ 1 & \frac{3 \pi }{4}\leq t<\frac{7 \pi }{4} \\ f(t-2 \pi\ \text{sgn}(t)) & \text{True} \\ \end{array} \right.\tag{1}$$
can be represented by the following shifted Fourier series:
$$f(t)=\underset{N\to\infty }{\text{lim}}\left(\sum_{n=1}^N a(n) \cos \left(n \left(t-\frac{\pi }{4}\right)\right)\right)\tag{2}$$
where
$$a(n)=\frac{2 \left(\sin \left(\frac{3 \pi n}{2}\right)-\sin \left(\frac{\pi n}{2}\right)\right)}{\pi n}\tag{3}$$
The following plot illustrates formula (2) for $f(t)$ in orange overlaid on the reference function defined in formula (1) in blue where formula (2) is evaluated using an upper evaluation limit of $N=16$.
Figure 1: Illustration of Formula (2) for $f(t)$
As another example the shifted square wave with peak-to-peak amplitude $2$ and period $2 \pi$ defined as follows
$$g(t)= \left\{ \begin{array}{cc} -1 & -\frac{3 \pi }{4}\leq t<\frac{\pi }{4} \\ 1 & \frac{\pi }{4}\leq t<\frac{5 \pi }{4} \\ g(t-2 \pi\ \text{sgn}(t)) & \text{True} \\ \end{array} \right.\tag{4}$$
can be represented by the following shifted Fourier series:
$$g(t)=\underset{N\to\infty }{\text{lim}}\left(\sum_{n=1}^N b(n) \sin\left(n \left(t-\frac{\pi }{4}\right)\right)\right)\tag{5}$$
where
$$b(n)=-\frac{2 \left((-1)^n-1\right)}{\pi n}\tag{6}$$
The following plot illustrates formula (5) for $g(t)$ in orange overlaid on the reference function defined in formula (4) in blue where formula (5) is evaluated using an upper evaluation limit of $N=16$.
Figure 2: Illustration of Formula (5) for $g(t)$