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Update: Solution for Example 1
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*It seems the right apparatus for studying such problems is *Malliavin calculus*, though I'm not at all yet familiar with the tool.*

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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][1], 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 $\Sigma = \sigma^2 I_d$).


  [1]: https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/kunze/malliavin/Malliavin_skript.pdf