Let $\tau>0$, $d\in\mathbb N$, $v:\mathbb R^d\to\mathbb R^d$ be Lipschitz continuous, $X^x\in C^0([-\tau,\tau],\mathbb R^d)$ be the solution of $$X^x(t)=x+\int_0^tv(X^x(s))\:{\rm d}s\;\;\;\text{for }t\in[-\tau,\tau]\tag1$$ for $x\in\mathbb R^d$, $$T_t(x):=X^x(t)\;\;\;\text{for }x\in\mathbb R^d$$ for $t\in[-\tau,\tau]$, $\Omega$ be a $d$-dimensional properly embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, $$\Omega_t:=T_t(\Omega)\;\;\;\text{for }t\in[-\tau,\tau],$$ $y_t:\Omega_t\to\mathbb R$ for $t\in[-\tau,\tau]$. Assume $${\rm d}y_0(\Omega;v):=\left.\frac{\rm d}{{\rm d}t}y_t(T_t(x))\right|_{t=0}\tag2$$ exists and $y_0$ is $C^1$-differentiable. Let $x\in\bigcap_{t\in[-\tau,\:\tau]}\Omega_t$.
How can we show that $$\left.\frac{\rm d}{{\rm d}t}y_t(x)\right|_{t=0}={\rm d}y_0(\Omega;v)-T_x(y_0)v(x)\tag3,$$ where $T_x(y_0)v(x)$ denotes the pushforward of $v(x)$ by $y(\Omega)$ at $x$.
I think the trick is to write $$y_t(x)-y_0(x)=y_t(x)-y_0(T^{-1}(x))-(y_0(x)-y_0(T^{-1}(x)))\tag4$$ and we may note that $$T_t^{-1}=T_{-t}\;\;\;\text{for all }t\in[0,\tau]\tag5,$$ but I don't know how to conclude.