Let $\tau>0$, $d\in\mathbb N$, $T_t$ be a $C^1$-diffeomorphism on $\mathbb R^d$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_{\mathbb R^d}$, $\Omega\subseteq\mathbb R^d$ and $y_t:\Omega_t:=T_t(\Omega)\to\mathbb R$ for $t\in[0,\tau)$.
Assume there is an open neighborhood $O$ of $\bigcup_{t\in[0,\:\tau)}\Omega_t$ and a Fréchet differentiable $Y:[0,\tau)\times O\to\mathbb R$ with $$\left.Y(t,\;\cdot\;)\right|_{\Omega_t}=y_t\;\;\;\text{for all }t\in[0,\tau).$$
Are we able to show that $\frac{\partial Y}{\partial t}(0,x)$ does not depend on the choice of $(\tau,O,Y)$ for all $x\in\Omega$?
Let $\tau=\infty$, $d=1$, $T_0=\operatorname{id}_{\mathbb{R}}$, $T_t\colon\mathbb{R}\rightarrow\mathbb{R},\,x\mapsto x+1$ for $t>0$, $\Omega=\{0\}\subset\mathbb{R}$, $\Omega_t:=T_t(\Omega)$ (i.e. $\Omega_0=\{0\}$ and $\Omega_t=\{1\}$ for $t>0$) and $y_t\colon\Omega_t\rightarrow\mathbb{R},\,x\mapsto0$. Then $\bigcup_{t\in[0,\infty)}\Omega_t=\{0,1\}$, so $O:=\mathbb{R}$ is an open neighborhood. The condition on $Y\colon[0,\infty)\times\mathbb{R}\rightarrow\mathbb{R}$ simply becomes $Y(0,0)=0$ and $Y(t,1)=0$ for $t>0$. This is satisfied by both $Y\colon[0,\infty)\times\mathbb{R}\rightarrow\mathbb{R},\,(t,x)\mapsto0$ and $Y^{\prime}\colon[0,\infty)\times\mathbb{R}\rightarrow\mathbb{R},\,(t,x)\mapsto(1-x)t$, both of which are differentiable (polynomial even), yet $\frac{\partial Y}{\partial t}(0,0)=0$ and $\frac{\partial Y^{\prime}}{\partial t}(0,0)=1$. So the choice does matter.
Most of the choices in this counter-example were arbitrary and a myriad others are probably possible. The one thing that makes this possible is choosing the diffeomorphisms so that $\Omega_0$ and $\Omega_t,\,t>0$ are separated. Knowing $\frac{\partial Y}{\partial t}(0,x),\,x\in\Omega_0$ requires knowing $Y(t,x)$ for $t$ in some neighborhood of $0$. We do know $Y(0,x)=y_0(x)$ by hypothesis, but we only know $Y(t,x)$ for $t>0$ directly if $x\in\Omega_t$, so making $\Omega_0$ and $\Omega_t,\,t>0$ disjoint removes this possibility. Making them separated just gives us enough space to freely interpolate between them, at which point there ought to be enough freedom for the partials to be just about anything.