Understanding equal power crossfades
Power of a (real-valued) digital signal $s[n]$ is simply the $s[n]^2$ (or proportional to that, depending if you want to normalize or not).

That's why the constant sum power constraint is $f^2(x) + f^2(1-x) = 1$. That's simply the definition of constant sum of  powers, nothing more.



so, what you've not specified is "logarithmic scale" to what, or what the perceived quantity is.

And the answer is perceived volume is logarithmic to power. The fact that this relation is logarithmic doesn't really matter here – it's only important that to achieve the same perceived volume, you need to have the same power.

Your functions $f$ work on amplitude, and thus, the square comes from the amplitude-to-power relation, not from the power-to-perceived-volume relation.


Ok, this question has become kind of a moving target. I'll try to explain what's up here, in the order my process of thought allows me to explain it:


Perceived volume is a function of signal energy. That is, to get the same perceived volume, you need to produce the same energy.
Power of a signal is proportional to the square of its amplitude, so with $s(t)$ being our signal, $s^2(t) =  \text{const.}$ restraint is very intuitive; energy of that  signal within $\tau$ time units is $\int_{T_0}^{T_0+\tau}s^2(t)\,dt$.
Your signal processing happens on amplitudes; you add amplitudes, not powers. Thus, let $s(t) = a(t) + b(t)$. It follows directly that $s^2(t) = a^2(t) + 2a(t)b(t) + b^2(t)$. Notice the middle $ab$ term!
But the formula we have dictates that $a^2(t) + b^2(t) = 1$. This only makes sense if $\int_{T_0}^{T_0+\tau} 2a(t)b(t)\,dt\equiv 0\quad\forall T_0, \tau$. Which is exactly the case if $a$ and $b$ are uncorrelated. If $a$ and $b$ were correlated, the $ab$ crosscorrelation term could not be equal to 0.