Let $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open, $$\mathcal A:=\{\Omega\subseteq U:\Omega\text{ is bounded and open and }\partial\Omega\text{ is of class }C^{0,\:1}\}$$ and $$\mathcal F(\Omega):=\int f:{\rm d}\sigma_{\partial\Omega)\;\;\;\text{for }\Omega\in\mathcal A$$ for some Borel measurable $f:U\to\mathbb R$ (what do we need to assume to ensure that $f$ is $\sigma_{\partial\Omega)$-integrable for all $\Omega\in\mathcal A$?

Given a family $(T_t)_{t\in[0,\:\tau)}$ of $C^1$-diffeomorphisms from $U$ onto $U$, where $\tau>0$, I want to consider the limit of $$\frac{\mathcal F(\Omega_t)-\mathcal F(\Omega_0)}t\tag1$$ as $t\to0$, where $\Omega\in\mathcal A$ and $\Omega_t:=T_t(\Omega)$ for $t\in[0,\tau)$.

The resulting expression will be the $\sigma_{\partial\Omega}$-integral of an expression involving the gradient of $f$. The question is: What kind of differentiability assumption of $f$ do we need to impose?

In general, $f$ would be called $C^1$-differentiable on $\partial\Omega$ if there is an $\mathbb R^d$-open neighborhood $O$ of $\partial\Omega$ and some $\tilde f\in C^1(O)$ with $$\left.f\right|_{\partial\Omega}=\left.\tilde f\right|_{\partial\Omega}\tag2.$$ I guess the problem with this is that it doesn't yield a well-defined notion of the derivative of $f$ on $\partial\Omega$, since $\nabla\tilde f(x)$ should depend on the choice of $\tilde f$ for $x\in\partial\Omega$.

I've seen frequently the assumption that $f\in C^1(T)$ for some tubular neighborhood $T$ of $\partial\Omega$, but I don't understand why this is a suitable (or even necessary) assumption. Why do we not simply assume that $f\in C^1(O)$ for some $\mathbb R^d$-open neighborhood $O$ of $\partial\Omega$ or even (which might be too restrictive, I guess) $f\in C^1(D)$?