Integral of the Laplace Beltrami operator multiplied by the Heaviside function

Let $$f \in W^{2,1}_p(U \times [0,T])$$ where $$U$$ is a $$C^2$$ compact manifold. I'm having a really hard time understanding why the following computation is valid:

$$\int_{U} H_\delta(f) \Delta_U f=-\int_U H_\delta'(f) {\vert \nabla f \vert }^2 (1)$$

where $$\Delta_U$$ denotes the Laplace Beltrami operator and $$H_\delta$$ stands for a smooth approximation of the Heaviside function such that $$\lim_{\delta \to 0^+} H_{\delta}=H$$ in the sense of distributions.

I know that in any compact manifold $$M$$ it holds (as a consequence of the divergence theorem):

$$\int_{M} f\Delta_M f=-\int_M {\vert \nabla f \vert }^2 (2)$$

Although it seems that $$(1),(2)$$ are strongly related, I can't conclude $$(1)$$ at the end! Probably I'm missing something about $$H_\delta$$

I'm really stuck here so I'd appreciate any help.

You need to use (assume that $$U$$ has no boundary, of $$f$$ vanishes at the boundary)
$$\int_U \nabla g \cdot \nabla f d\mu = \int g \Delta f d\mu$$
with the chain rule $$\nabla (H_\delta (f)) = H_\delta '(f) \nabla f$$.