Let $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open, $\tau>0$, $T_t$ be a $C^1$-diffeomorphism from $U$ onto $U$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_U$.
Let $V_2\subseteq\mathbb R^d$ be open with $V_2\subseteq U$. Can we show that there is an open $V_1\subseteq\mathbb R^d$ and a $\delta>0$ with $$T_t(V_1)\subseteq V_2\;\;\;\text{for all }t\in[0,\delta)?\tag1$$
We clearly need to assume some kind of regularity (continuity) of $$[0,\tau)\times U\ni(t,x)\mapsto T_t(x)\tag2.$$ Under a suitable assumption, the claim should hold, since $T_0=\operatorname{id}_U$. So, what do we need to assume?
Assume the continuity of the map $(t,x)\mapsto T_t(x)$. Since $V_2$ is an open set of $\mathbb{R}^d$, then $T^{-1}(V_2)$ is an open set of $I\times \mathbb{R}^d$ ($I=[0,\tau)$), which contains $\{0\}\times V_2$ since $T_0(x)=x$ for all $x\in V_2$. From now on, choose an arbitrary $y\in V_2$.
Since the set $I\times\mathbb{R}^d$ is endowed with the product topology, there are open sets $J\subset I$ and $V_1\in\mathbb{R}^d$ such that $0\in J,y\in V_1$ and $J\times V_1\subset T^{-1}(V_2)$. You can restrict $J$ to an interval of the form $[0,\delta)$, thus getting the $\delta$ and $V_1$ you were looking for.