of $X'^2$ can be computed using the chain rule - $(2\psi_{0,2}(\sqrt{x}) / (2\sqrt{x})=\frac{1}{\sqrt{4\pi }} e^{-x/4} x^{-1/2} = \gamma_{1/4, 1/2} (x),$ where $\gamma $ is a gamma density. Now it's an exercise (use convolutions or change of variables) to show that the sum of $\Gamma_{\alpha_1 , \beta }, \Gamma_{\alpha_2, \beta }$ distributed