Let $d\in\mathbb N$. I want to compute the shape derivative of a shape functional$^1$ $$\mathcal F(\Omega):=\int_{\partial\Omega}f\:{\rm d}\sigma_{\partial\Omega}\;\;\;\text{for }\Omega\in\mathcal A$$ for some suitably integrable real-valued function $f$ on a suitable large subspace of $\mathbb R^d$ and $$\mathcal A:=\{\Omega\subseteq\mathbb R^d:\Omega\text{ is bounded and open and }\partial\Omega\text{ is of class }C^1\}.$$ If $\tau>0$ and $T_t$ is a $C^1$-diffeomorphism on $\mathbb R^d$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_{\mathbb R^d}$, $\Omega\in\mathcal A$ and $$\Omega_t:=T_t(\Omega)\;\;\;\text{for }t\in[0,\tau),$$ then we can show that $$\mathcal F(\Omega_t)=\int_{\partial\Omega}\left|\det{\rm D}T_t(x)\right|\left\|({\rm D}T_t(x)^{-1})^\ast\nu_{\partial\Omega}(x)\right\|(f\circ T_t)(x)\:\sigma_{\partial\Omega}({\rm d}x)\tag1$$ for all $t\in[0,\tau).$ Since $T_0=\operatorname{id}_{\mathbb R^d}$, we can assume that $\tau$ is small enough to ensure $\det{\rm D}T_t(x)>0$ for all $(t,x)\in[0,\tau)\times\mathbb R^d$.
Ignoring rigor, we easily see that $$\frac{\mathcal F(\Omega_t)-\mathcal F(\Omega_0)}t\xrightarrow{t\to0+}\int_{\partial\Omega}(\nabla_{\partial\Omega}\cdot v_0)(x)f(x)\langle\nabla f(x),v_0(x)\rangle\:\sigma_{\partial\Omega}({\rm d}x)\tag2,$$ where $$(\nabla_{\partial\Omega}\cdot v_0)(x)=(\nabla\cdot v_0)(x)-\langle{\rm D}v_0(x)\nu_{\partial\Omega}(x),\nu_{\partial\Omega}(x)\rangle\;\;\;\text{for all }x\in\partial M.$$
My question is: Which assumptions on $f$ do we need to impose? Since all $\Omega\in\mathcal A$ are bounded, all the measures $\sigma_{\partial\Omega}$ are finite. So, it would be sufficient to assume that $f$ is a bounded function on all of $\mathbb R^d$. However, this assumption might be too restrictive. And whatever we come up with, we need to assume that $f$ is continuously differentiable on the set it is defined and the derivative must be integrable as well.
If we consider only the single fixed $\Omega$, I've read that it's sufficient to assume $f\in C^1(U)$ for some tubular neighborhood of $\partial\Omega$. How can we show that? Why is this a suitable assumption?
$^1$ $\sigma_{\partial\Omega}$ denotes the surface measure on the Borel $\sigma$-algebra $\mathcal B(\partial M)$.