Why is the constant $c$ replaced by velocity $v$ in this particular derivation? 
Is this derivation for de Broglie's wavelength correct?
If yes, why is it that we're changing $c$ to $v$? Is that allowed? 
If we use that logic and reverse engineer the whole derivation, we get $E=mv^2$ for $v<c$ and not $E=mc^2$. Isn't $E=mv^2$ only true for $v=c$ and not $v<c$?
 A: For massive particles, the de Broglie relations
$$
f = \frac Eh
\qquad\qquad
\lambda = \frac hp
$$
or equivalently
$$
E = \hbar\omega
\qquad\qquad
\vec p = \hbar \vec k
$$
are not derived, but postulated.
For photons, they were discovered experimentally (photoelectric effect, Compton scattering, ...). De Broglie then hypothesized that these equations held universally, which was eventually confirmed experimentally through diffraction of electron beams.
Once you assume these relations, what you can do is for example derive the phase and group velocity of matter waves: Group velocity will agree with ordinary particle velocity, whereas phase velocity will be $v_p = c^2/v$.
In contrast, what you're doing is throwing together relations that only hold for massless particles with those that hold for massive ones. There's also a factor of $\gamma$ missing.
Properly, it needs to read
$$
E = \gamma mc^2 = hf
$$
where
$$
f = \frac{v_p}{\lambda} = \frac{c^2}{\lambda v}
$$
hence
$$
\gamma mc^2 = \frac{hc^2}{\lambda v}
\implies \lambda = \frac h{\gamma mv} = \frac hp
$$
This, however, already assumed the de Broglie relations in the derivation of $v_p$.
