How to filter a signal using a bandpass filter consisted of two moving average filters?
The Japanese link actually implies how to derive Eq.5, and ironically they also refer to an existing dsp.se answer at the bottom.
Derivation of Eq.5 is as follows:
Consider a moving average filter of length $N$, with the impulse response $h[n]$:
$$h[n] = \begin{cases} ~~~1/N~~~,~~~n=0,1,...,N-1 \\ ~~~0~~~,~~~ \text{otherwise} \end{cases} \tag{1}$$
Magnitude of its frequency response $~H(\omega)~$  (DTFT of $h[n]$) can be shown to be:
$$ |H(\omega)| = \frac{1}{N} \left|\frac{ \sin(\frac{\omega}{2}N)}{\sin(\frac{\omega}{2})}\right| \tag{2}$$
Now, after replacing the $\sin()$ functions with their Taylor expansions in powers of $\omega$, it can be shown by polynomial long division that Eq.2 is also given by :
$$|H(\omega)| = 1 + \frac{1}{24}(1-N^2) \omega^2 + H.O.T. \tag{3}$$ where H.O.T. refers to higher order terms in powers of $\omega$. An approximation for small values of $\omega$ is obtained by neglecting H.O.T.:
$$|H(\omega)| \approx 1 + \frac{1}{24}(1-N^2) \omega^2 \tag{4}$$
Using this approximate frequency response magnitude, we can obtain an approximate cutoff frequency $\omega_c$ at which the magnitude $|H(\omega_c)|$ falls to $1/\sqrt{2}$ of its value at $\omega = 0$, which is $H(0) = 1$.
$$ |H(\omega_c)| = \frac{1}{\sqrt{2}} = 1 + \frac{1}{24}(1-N^2) \omega_c^2 \tag{5}$$
Replace the discrete-time frequency $w_c$ by $w_c = 2\pi f_c /f_s$, where $f_c$ is the analog cutoff frequency in Hz, and $f_s$ is the sampling frequency in Hz. Finally solving the resulting algebraic expression for $f_c$ yields the formula that you refer to as Eq.5 in the document :
$$ |H(\omega_c)| = \frac{1}{\sqrt{2}} = 1 + \frac{1}{24}(1-N^2) \left( 2\pi \frac{f_c}{f_s} \right)^2 \tag{6}\\\\$$
$$ f_c = \frac{1}{\pi} \frac{\sqrt{6 - 3\sqrt{2}}}{\sqrt{N^2-1}} ~f_s ~~ =~~  \frac{0.422}{\sqrt{N^2-1}} ~f_s \tag{7} $$
Eq.7 above is the formula that provides an approximate cutoff frequency calculation for the moving average filter of length $N$ (order $N-1$). In your posted link, Eq.5 there's a slight variation at the scale of $0.442$ instead of $0.422$, probably they tried to apply some correction to the actual cutoff vs approximated one.
Note that in the derivation, we've used an approximation of the DTFT magnitude which was valid as long as $\omega$ was small compared to $\pi$. This means that the approximation will be satisfactory if $\omega_c$ is close to $0$, or in other words, $f_c$ is a small compared to $f_s$. And indeed this will be the case for high order moving average filters. And the approximation gets better as $N$ increases.
Using such two moving average filters with approximate cutoff frequencies $f_{c1}$ and $f_{c2}$ to create a bandpass filter will not be very satisfactory unless your out of band signal energy is insignificand after 20 to 30 dB of attenuation.