$\newcommand\R{\mathbb R}$Let $f$ be a radial pdf on $\R^n$, so that $$f(x)=g(|x|)\tag{1}\label{1}$$ for some function $g\colon[0,\infty)\to\R$ and all $x\in\R^n$, where $|x|:=|x|_2$. Then your desired respresentation $$f(x)=\int_0^\infty h(w) \cdot 1_{w \mathbb B^n} \ dw\tag{2}\label{2}$$ with $h\ge0$ can be rewritten as $$g(|x|)=\int_0^\infty dw\,h(w)\,1(|x|\le w)$$ for all $x\in\R^n$ and then further as $$g(u)=\int_u^\infty dw\,h(w)$$ for all real $u\ge0$, with $h\ge0$.
So, your desired representation \eqref{2} for $f$ as in \eqref{1} holds if and only if $g$ is a nonincreasing absolutely continuous function such that $g(u)\to0$ as $u\to\infty$. Moreover, then $h=-g'$ almost everywhere.
The formula $$f(\mathbf x) = \frac{1}{(2\pi)^{n/2}}\int_{w=0}^\infty \exp\left(-w^2\right) \cdot 1_{w\mathbb S^{n-1}}(\mathbf x) \ dw\tag{3}\label{3}$$ for $ f(\mathbf x) = \frac{1}{(2\pi)^{n/2}} \exp\left(-|\mathbf x|_2^2\right)$ in the very beginning of your argument is incorrect. Indeed, the integral in \eqref{3} is in fact $$\int_0^\infty \exp\left(-w^2\right) \,1(w=|\mathbf x|_2) \, dw=0\ne f(\mathbf x).$$