Let $d\in\mathbb N$, $\mathcal M$ denote the set of bounded $d$-dimensional properly embedded $C^1$-submanifolds of $\mathbb R^d$ with boundary, $f\in C^1(\mathbb R^d)$, $$\mathcal G(\partial M):=\int_{\partial M}f\:{\rm d}\sigma_{\partial M}\;\;\;\text{for }M\in\mathcal M,$$ $\tau>0$ and $v:[0,\tau]\times\mathbb R\to\mathbb R$ be continuous in the first argument with $v(t\;\cdot\;)\in C^1(\mathbb R^d,\mathbb R^d)$ for all $t\in[0,\tau]$ and ${\rm D}_2v\in C^0([0,\tau]\times\mathbb R^d,\mathfrak L(\mathbb R^d))$. We know that, for all $x\in\mathbb R^d$, there is an unique $X^x\in C^0([0,\tau])$ with $$T_t(x):=X^x(t)=x+\int_0^tv(s,X^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]\tag1.$$ For simplicity, write $v_0:=v(0,\;\cdot\;)$.
If $M\in\mathcal M$, we can show that $$\tag2\left.\frac{\rm d}{{\rm d}t}\mathcal G(\partial T_t(M))\right|_{t=0+}=\int f(x)(\operatorname{div}_{\partial M}v_0)(x)+\langle\nabla f(x),v_0(x)\rangle\:\sigma_{\partial M}({\rm d}x),$$ where $$(\operatorname{div}_{\partial M}v_0)(x):=\operatorname{tr}{\rm D}v_0(x)\operatorname P_{\partial M}(x)$$ and $\operatorname P_{\partial M}(x)$ denotes the orthogonal projection of $\mathbb R^d$ onto $T_x\:\partial M$ for $x\in\partial M$.
Assume $$v(t,x)\in T_x\:\partial M\;\;\;\text{for all }(t,x)\in[0,\tau]\times\partial M.\tag4$$ The intuition tells us that $(2)$ should be equal to $0$, since the velocity evolves tangential to the boundary. And, in fact, if we assume that $M$ is even $C^2$-regular, $(2)$ is equal to $$\int\left((\operatorname{div}\nu_{\partial M})(x)f(x)+\frac{\partial f}{\partial\nu_{\partial M}}(x)\right)\underbrace{\langle v_0(x),\nu_{\partial M}(x)\rangle}_{=\:0}\:\sigma_{\partial M}({\rm d}x)=0\tag5,$$ where $\nu_{\partial M}$ denotes the outward pointing unit normal field on $\partial M$.
How can we see that $(2)$ is equal to $0$ without assuming the $C^2$-regularity?