First, let us prove the following general result:
Proposition. Let $M$ be a manifold and $p\in M$, let $X$ be a vector field on $M$ and let $g\colon M\rightarrow\mathbb{R}$ smooth. If $X(p)\neq 0$, then there exists $U$ an open neighborhood of $p$ in $M$ and $f\colon U\rightarrow\mathbb{R}$ smooth such that:
$$\mathrm{d}f(X)=g_{\vert U}.$$
Furthermore, if $g(p)=0$, one may assume that $f(p)=0$ and $\mathrm{d}f_p=0$.
Proof. Using the straightening theorem, there exists $(U,\phi)$ a chart of $M$ around $p$ such that $\phi_*X=\frac{\partial}{\partial x_1}$. Furthermore, one may assume that for $(x_1,\ldots,x_n)\in\phi(U)$, one has the following property:
$$s\in[\min(0,x_1),\max(0,x_1)]\Rightarrow(s,x_2,\ldots,x_n)\in\phi(U)\tag{$\star$}.$$
Therefore, one can define a smooth map $F\colon\phi(U)\rightarrow\mathbb{R}$ by the following formula:
$$F(x_1,\ldots,x_n):=\int_{0}^{x_1}g(\phi^{-1}(s,x_2,\ldots,x_n))\,\mathrm{d}s.$$
Notice that one has $F(0)=0$ and $\frac{\partial F}{\partial x_1}=g\circ\phi^{-1}$.
In this section, assume that $g(p)=0$, then there exists constants $a_2,\ldots,a_n$ such that $\mathrm{d}F_0=\sum\limits_{i=2}^na_i\mathrm{d}x_i$
and let us define $\overline{F}\colon\phi(U)\rightarrow\mathbb{R}$ in the following fashion:
$$\widetilde{F}(x_1,\ldots,x_n):=F(x_1,\ldots,x_n)-\sum_{i=2}^na_ix_i.$$
Notice that $\widetilde{F}(0)=0$, $d\widetilde{F}_0=0$ and $\frac{\partial\widetilde{F}}{\partial x_1}=g\circ\phi^{-1}$, so that one can assume that $F(0)=0$ and $\mathrm{d}F_0=0$.
Finally, with $f=F\circ\phi$, using the chain rule, for all $x\in U$, one has:
$$\mathrm{d}f_x(X(x))=(\mathrm{d}F_{\phi(x)}\circ T_x\phi)(X(x))=\mathrm{d}F_{\phi(x)}\left(\frac{\partial}{\partial x_1}_{\big\vert\phi(x)}\right)=\frac{\partial F}{\partial x_1}_{\big\vert\phi(x)}=g(x).$$
In addition, if $g(p)=0$, then one has $\mathrm{d}f_p=\mathrm{d}F_0\circ T_p\phi=0$. Whence the result. $\Box$
Remark. When I say that $(U,\phi)$ a chart aroud $p$, I mean $\phi(p)=0$.
Remark. Let $\|\cdot\|$ be the product norm on $\mathbb{R}^n$, since $0\in\phi(U)$ is open, there exists $\varepsilon>0$, s.t. $B(0,\varepsilon)\subset U$.
Let $x\in B(0,\varepsilon)$ and $s\in[\min(0,x_1),\max(0,x_1)]$, then notice that $|s|\leqslant|x_1|$, so that one has:
$$(s,x_2,\ldots,x_n)\in B(0,\varepsilon).$$
Therefore, shrinking $U$ to $\phi^{-1}(B(0,\varepsilon))$ establishes the technical assumption $(\star)$.
For all $t\in[0,1]$, since $R_{\alpha_t}(0)\neq 0$ (for example: $R_{\alpha_t}(0)\not\in\xi_0$) applying the result to the map $-\dot{\alpha_t}(R_{\alpha_t})$, there exists $U_t$ an open neighborhood of $0$ in $\mathbb{R}^{2n+1}$ and a map $H_t\colon U_t\rightarrow\mathbb{R}$ such that:
$$\dot{\alpha_t}(R_{\alpha_t})+\mathrm{d}H_t(R_{\alpha_t})=0.$$
To conclude, one has to see that there exists a single neighborhood on which all $H_t$, $t\in[0,1]$ are defined. For all $t\in[0,1]$, let us define the following time of existence:
$$\varepsilon_t:=\sup\{\varepsilon>0\textrm{ s.t. }B(0,\varepsilon)\subset U_t\},$$
then $t\mapsto\varepsilon_t$ is a lower semicontinuous function and by compactness of $[0,1]$, one has $\varepsilon:=\inf\limits_{t\in[0,1]}\varepsilon_t>0$. Finally, for all $t\in[0,1]$, $H_t$ is defined on $B(0,\varepsilon)$.
