I currently have a standard result regarding variances and it looks like:
$$\frac{(2n-1)s^2}{\sigma^2} \sim \chi^2_{2n-1}$$
that is, it is approximately a chi-squared distribution with $2n-1$ degrees of freedom.
Now, can I take the variance of both sides so that I get:
$$\frac{(2n-1)^2 \operatorname{var}(s^2)}{\sigma^4} = 2(2n-1)$$
(since the variance of chi-square is just the degrees of freedom times $2$).
Would this be a valid step? Thank you!
The symbol $\sim$ does not mean "approximately" in this context (use $\approx$ instead, for "equals approximately"). It means "follows exactly the distribution of a..." or other verbal transcription to that effect.
So if we assume that
$$\frac{(2n-1)s^2}{\sigma^2} =Q \sim \chi^2_{2n-1}$$
we have that the random variable $Q$ follows a chi-square with $2n-1$ degrees of freedom.
It is then perfectly valid to write
$$s^2 =\frac{\sigma^2}{2n-1}Q$$
This makes the random variable $s^2$ to follow a Gamma distribution,
$$s^2 \sim \Gamma_{d}\left(\frac {2n-1}{2},2\frac{\sigma^2}{2n-1}\right)$$ where we have used the "shape-scale" parametrization. Then
$$\text{Var}(s^2) = \frac {2n-1}{2}\left(2\frac{\sigma^2}{2n-1}\right)^2 = \frac {2\sigma^4}{2n-1}$$
which is what you indeed found -but it is advisable to go through the above procedure, specifically, to use the equality symbol together with a variable symbol (like the $Q$ I used), before performing mathematical manipulations.
If on the other hand what we assume is that $\frac{(2n-1)s^2}{\sigma^2}$ follows a chi-square distribution only "approximately",
$$\frac{(2n-1)s^2}{\sigma^2} \approx Q \sim \chi^2_{2n-1}$$
then still, the above calculations are not invalid, but, the accuracy of the obtained expression for the variance of $s^2$ should be under questioning and investigation (since $=$ should be everywhere changed to $\approx$).
