after reviewing the properties of the Dirac delta function, I have a hard time figuring out one property: the composition rule.
I understand the integral form comes from a change of variable and by partitioning the domain around the roots of a function g, but I don't understand the first form.
Assume that $f \in C^\infty(\mathbb{R})$ satisfies $f(x_i) = 0$ with $f'(x_i) > 0.$ Then there is some interval $(x_i-\epsilon, x_i+\epsilon)$ on which $f'>0$ and $f$ is invertible. Then, using the substituion $y = f(x),\ dy = f'(x) \, dx,$ $$\begin{align} \int_{x_i-\epsilon}^{x_i+\epsilon} \delta(f(x)) \, \varphi(x) \, dx = \int_{f(x_i-\epsilon)}^{f(x_i+\epsilon)} \delta(y) \, \varphi(f^{-1}(y)) \, \frac{dy}{f'(f^{-1}(y))} \\ = \varphi(f^{-1}(0)) \, \frac{1}{f'(f^{-1}(0))} = \varphi(x_i) \, \frac{1}{f'(x_i)} \\ = \frac{1}{f'(x_i)} \int_{x_i-\epsilon}^{x_i+\epsilon} \delta(x-x_i)\,\varphi(x)\,dx\\ = \int_{x_i-\epsilon}^{x_i+\epsilon} \frac{\delta(x-x_i)}{f'(x_i)} \,\varphi(x)\,dx \end{align}$$
If instead $f'(x_i) < 0,$ then $f(x_i-\epsilon) > f(x_i+\epsilon)$ and we get $$\begin{align} \int_{x_i-\epsilon}^{x_i+\epsilon} \delta(f(x)) \, \varphi(x) \, dx = \int_{f(x_i-\epsilon)}^{f(x_i+\epsilon)} \delta(y) \, \varphi(f^{-1}(y)) \, \frac{dy}{f'(f^{-1}(y))} \\ = \{ \text{ swapping integration limits } \} \\ = - \int^{f(x_i-\epsilon)}_{f(x_i+\epsilon)} \delta(y) \, \varphi(f^{-1}(y)) \, \frac{dy}{f'(f^{-1}(y))} \\ = - \varphi(f^{-1}(0)) \, \frac{1}{f'(f^{-1}(0))} = - \varphi(x_i) \, \frac{1}{f'(x_i)} \\ = - \frac{1}{f'(x_i)} \int_{x_i-\epsilon}^{x_i+\epsilon} \delta(x-x_i)\,\varphi(x)\,dx\\ = - \int_{x_i-\epsilon}^{x_i+\epsilon} \frac{\delta(x-x_i)}{f'(x_i)} \,\varphi(x)\,dx \end{align}$$
The two cases can be combined into $$ \int_{x_i-\epsilon}^{x_i+\epsilon} \delta(f(x)) \, \varphi(x) \, dx = \int_{x_i-\epsilon}^{x_i+\epsilon} \frac{\delta(x-x_i)}{|f'(x_i)|} \,\varphi(x)\,dx $$
Taking all zeroes into account we get $$ \int_{-\infty}^{\infty} \delta(f(x)) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} \,\varphi(x)\,dx , $$ i.e. $$ \delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{|f'(x_i)|} . $$