Saw this particular statement from my lectures
$\frac{(n-p)s^2}{\sigma^2} \sim \chi^2$ then $(n-p)s^2 \sim \sigma^2 \chi^2$
My first question was what does $\sigma \chi^2$ even mean?
Secondly, why are we able to just multiplying both sides by $\sigma^2?$ Are we just treating $\sim $ like an $=$ sign? If so, why?
I think your confusion arises from what can be considered an abuse of notation in the equation you quote. If you treat $ \chi^2 $ as a random variable, then the relation makes sense because for any two random variables $X,Y$ on the same probability space $\left(\Omega,\mathcal{F},P\right)$ if $X\sim Y$ then $f(X)\sim f(Y)$ for any Borel-measurable function $f$. To see this, let $S \in \mathcal{B}\left(\mathbb{R}\right)$, that is, let $S$ be a Borel set in $\mathbb{R}$ and observe that $$P(f(X)\in S) = P(X \in f^{-1}(S)) = P(Y \in f^{-1}(S)) = P(f(Y)\in S)$$ where the second equality follows from the fact that $X,Y$ are identically distributed. While I am being precise and using the language of measure theory, the basic idea is very simple and the measurability condition is usually satisfied by most functions. In particular, it is satisfied by $f(x) = \sigma x$