Update: Solution for Example 1
It seems the right apparatus for studying such problems is Malliavin calculus, though I'm not at all yet familiar with the tool.
So, let $G = (g_1,\ldots,g_d)$ be a standard Gaussian random vector in $\mathbb R^d$ and consider the Gaussian process defined by $Z(a) := \langle G,a\rangle = \sum_j a_j g_j$, for any $a \in \mathbb R^d$. The random variable "$f(x)$, $x \sim N(m,\Sigma)$" in the question can be written as
$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.
Let $S(X):=\delta(DX/\|DX\|^2)$, where $D$ (resp. $\delta$) is the Mallavian derivative (resp. Skorohod integral) operator. A simply computation gives $DX = \widetilde w$, and so $$ S(X) = (1/\|\widetilde w\|^2)\sum_j \widetilde w_j G_j = (1/\|\widetilde w\|)Z(\widetilde w/\|\widetilde w\|) $$
Therefore, thanks to Proposition 2.1.1 of Introduction to Malliavin Calculus, one obtains
$$ \begin{split} s_f'(t) &= \mathbb E[1_{X \le t} S(X)] = \frac{1}{\|\widetilde w\|}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{(t-f(m))/\|\widetilde w\|}ze^{-z^2/2}dz = \varphi(\frac{t-f(m)}{\|\widetilde w\|})\\ &= \varphi(\frac{t-f(m)}{\|w\|_\Sigma}) \end{split} $$ where $\varphi$ is the standard Gaussian pdf. This exactly matches the computation carried out in the comments section to the question above (where $\Sigma = \sigma^2 I_d$).