I have a little knowledge of statistics, so correct me if the question is not well-phrased/formulated.
Let $\{X_1, \dots, X_N\}$ be independent normal random variables with mean 0 and variance $\sigma^2$. If we define $Z$ as $Z = \frac{\sigma^2}{N}\sum_{i=1}^N|X_i^2|$, then $\sum_{i=1}^N|X_i^2|$ is a chi-squared random variable with N degrees of freedom. Now assume for the same sample set, each sample $X_i$ is a normal random variable with mean 0 and: variance $\sigma^2 = \sigma_1^2$ with probability $p$; and with variance $\sigma^2 = \sigma_1^2 + \sigma_2^2$ with probability $q = 1 - p$.
If I define the same $Z = \frac{\sigma^2}{N}\sum_{i=1}^N|X_i^2|$, then what is the degree of freedom of $\sum_{i=1}^N|X_i^2|$? and why? I assume it should be $N$ because the sample size is $N$, but what I see in the papers is $2N$. Thanks in advance.
In your original same-variance example, $\frac{1}{\sigma^2}\sum_{i=1}^NX_i^2$ (you don't need the modulus signs) is $\chi_N^2$-distributed, with mean $N$ determining the coefficient. In your second example, note $X_i^2$ has mean equal to the mean variance of $X_i$, i.e. $p\sigma_1^2+(1-p)(\sigma_1^2+\sigma_2^2)=\sigma_1^2+q\sigma_2^2$. But it does't have a $\chi^2$ distribution, even up to a scaling; nor does $Z$. One way to prove this is with moment-generating functions. The MGF of $X_i^2$ is $p/(1-2i\sigma_1t)+q/\left(1-2i\sqrt{\sigma_1^2+\sigma_2^2}t\right)$, which isn't of the form $1/(1-2i\sigma t)^\nu$.