I am confused with the derivation of the Rice's distribution. The resulting density of the sum of two squared normal random variables is the chi-squared density. Yet, when deriving the Rice's distribution as, $$ R= \sqrt{X_1^{2} + X_2^{2}} $$ that is, transforming a chi-squared distribution using $$ f(r) = \sqrt{u} $$ where $$ u \sim \chi $$ does not arrive to the known Rice's distribution. Indeed, the derivation makes use of the multiplication of two independent random variables, and then, transforming them from Cartesian to polar.
The proof I see is shown in the link below, and it seems it is the usual proof. ftp://ftp.bartol.udel.edu/anita/amir/Rician_Distribution_notes.pdf
You actually can use a variable transformation; you're just transforming the wrong distribution. Let $f(z) = \sqrt{x}$ where $x$ is a random variable sampled from a chi-square distribution with two-degrees of freedom. The pdf of $x$ is:
$$p_x(x) = \frac{1}{2\sigma^2} \exp{\left(\frac{-x}{2\sigma^2}\right)} \;\; (x\geq 0)$$ Simplifying the steps of the transformation, the pdf of $f(z)$ is: $$p_z(z) = \frac{d}{dz}\left[\int_0^{z^2}p_x(x)dx\right] = 2|z| \cdot p_x(z^2) = \frac{z}{\sigma^2} \exp{\left(\frac{-z^2}{2\sigma^2}\right)}, \;\; (z\geq 0) $$ which is the Rayleigh distribution, not the Rician.
In order to arrive at the Rician we have to consider Gaussians with non-zero means. Let's instead transform a non-central chi-square distribution (with two-degrees of freedom). The pdf of $x$ is now:
$$p_x(x) = \frac{1}{2\sigma^2} \exp{\left(-\frac{x+\nu^2}{2\sigma^2}\right)} I_0\left(\frac{\nu\sqrt{x}}{\sigma^2} \right) \;\; (x\geq 0)$$
where $\nu^2 = \mu_1^2 + \mu_2^2$ is the magnitude square of the two means and $I_0$ is the zero'th-order modified Bessel function of the first kind. Again, letting $f(z) = \sqrt{x}$ we can repeat the derivation and find the pdf of $f(z)$ to be:
$$p_z(z) = \frac{d}{dz}\left[\int_0^{z^2}p_x(x)dx\right] = 2|z| \cdot p_x(z^2) = \frac{z}{\sigma^2} \exp{\left(-\frac{z^2+\nu^2}{2\sigma^2}\right)} I_0\left(\frac{\nu z}{\sigma^2} \right), \;\; (z\geq 0) $$ Which is the desired Rician Distribution.