I have the same reservations as you about the derivation that you found. I prefer this treatment ...
For a photon
$$E=hf=\frac{hc}\lambda.$$
We can show, using classical electromagnetic theory, that a 'slice' of an electromagnetic wave carrying energy $E$ in its fields also carries momentum $E/c$. Presumably a photon's momentum, $p$, is related to its energy in the same way. [We can reach the same conclusion by taking a photon to be a particle moving at speed $c$, and therefore massless, for which we can put $m=0$ in $E^2-c^2p^2=c^4m^2$.]
So...
$$p=\frac Ec\ \ \ \ \ \ \text {therefore}\ \ \ \ \ \ p=\frac h\lambda.$$
De Broglie famously suggested that the same relationship applies for a particle of matter, which, he supposed, has a wave aspect to it. This was an hypothesis rather than a logical deduction from established Physics, though de Broglie did note fragments of support for it, such as the light it seems to shed on Bohr's condition for a stable electron orbit in an atom: $pr=nh/2\pi$... If an electron wave fits without discontinuities round a circle of radius centred on the nucleus, then $n\lambda=2\pi r$. Substituting $h/p$ for $\lambda$ gives the Bohr condition!
We'll now look in more detail at the de Broglie wave for a particle moving at constant velocity, $\mathbf u$. De Broglie assumed that the Planck relationship between energy and frequency applies to the particle, that is
$$\ \ \ \ \ \ \ \ \ E=hf\ \ \ \ \ \ \ \text{so}\ \ \ \ \ \ \ \gamma m c^2=hf\ \ \ \ \ \ \ \ \ \text{in which}\ \ \ \ \ \ \ \ \ \ \gamma=\frac1{\sqrt{1-u^2/c^2}},$$
as well as
$$p=\frac h \lambda\ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \gamma mu=\frac h{\lambda}$$
Dividing $\gamma m c^2=hf$ by $\gamma mu=\frac h{\lambda}$,
$$v=\frac {c^2} u$$
in which $v=f \lambda$ and represents then phase speed of the de Broglie wave.
At first sight this is upsetting: not only is $v$ different from $u$, but since $u<c$ we have $v>c$. But now we note that at any time a particle has a more or less well defined position in space, whereas a wave of sharply defined frequency has an infinite extent. But many such waves of not quite the same frequency and wavelength as each other can interfere to form a wave packet of finite extent. The wave energy travels, not at the phase speed $v$ of a constituent wave, but at the speed of the packet, the so-called group speed, $v_{\text{group}}$. It is this group speed that must be less than $c$.
To calculate the group speed neatly, we use wave variables $\omega$ and $k$, defined by
$$\omega= 2\pi f\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ k=\frac{2\pi}\lambda$$
Thus, phase speed, $v$, can be written as
$$v=f \lambda=\frac{\omega}k$$
We can show that for any wave packet...
$$v_{\text{group}} =\frac{d\omega}{dk}$$
With our new notation, and writing $\hbar=\frac h{2\pi}$, the Planck and de Broglie relationships become
$$E=\hbar \omega\ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ p=\hbar k.$$
Special relativity gives us
$$E^2-c^2 p^2=c^4 m^2\ \ \ \ \ \ \ \text{so}\ \ \ \ \ \ \ \hbar^2 \omega^2-c^2 \hbar^2 k^2=c^4 m^2$$
From which we deduce that
$$\frac{d\omega}{dk}=c^2 \frac k \omega\ \ \ \ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \ \ \ v_{\text{group}}=\frac{c^2}v=u$$
The group speed is therefore equal to the particle speed, a satisfying result. de Broglie referred to the particle as having a pilot wave. He did not subscribe to a probability interpretation of the wave.
