Discontinuous pdf of 2 branch Random Variable Transformation

Given the pdf of a random variable $X$ $$f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$ I want to find the pdf of random variable $Y$ defined: $$ Y = g(X) = \left\{ \begin{array}{ll} 4X & X \geq 0\\ -2X & X < 0 \end{array} \right. $$ After working out the problem, I found $$f_Y(y) = \frac{1}{4\sqrt{2\pi}}e^{-y^2/32} + \frac{1}{2\sqrt{2\pi}}e^{-y^2/8} , y > 0$$ Y does not take negative values so we can say that $\{Y < 0\} = \emptyset $ and thus $f_Y(y) = 0$ if $y < 0$.

However, what can I say about $f_Y(0)$, since the derivative of the transformation is not defined at $X = 0$. I know that in the integral of $f_Y(y)$ we don't care about this value since we have in any case a bounded discontinuity. But is there a "more correct" value for $f_Y(0)$?

The short answer is it doesn't matter what value you assign to the pdf at $y=0$. In fact, it doesn't really matter what value is assigned to the pdf at any countable collection of points in this case (or any measure zero set more generally), since such modifications don't change the underlying distribution (i.e. CDF) of $Y$.

At the end of the day, a pdf is a convenient object that we integrate out to compute probabilities.