I am not sure if I understand the definition of a contact vector field:
Let $(W, \xi)$ be a contact manifold. A vector field $Y$ is called a contact vector field, if the flow of $Y$ preserves $\xi$.
If $\lambda$ is a defining contact form, then this means that $\mathcal{L}_Y \lambda= p \lambda$ where $p: W \rightarrow \mathbb{R}$.
Now what exactly does "the flow of $Y$ preserves $\xi$" mean?
As far as I understand, it is the following:
Let $\varphi_t(x)$ be the flow of $Y$. Then for each $t$, it holds:
$\varphi_t^{*} \xi = \xi $ so for each tangent vector $v \in \xi_p$, it holds that the tangent vector
$f \mapsto v(f \circ \varphi_t)$ is in $\xi_{\varphi_t(p)}$
Is that correct?
And if yes, how does it correspond to
$\mathcal{L}_Y \lambda= p \lambda$ where $p: W \rightarrow \mathbb{R}$
Locally, $\xi= \ker\{\lambda\}$. $\mathcal{L}_Y \lambda = d(\iota_Y \lambda)+ \iota_Y(d \lambda)$
But why is that equivalent?
Let $X$ be a contact vector field, and $\phi_t$ its flow, we have for every $x\in W, u\in T_x\xi$, $(\phi_t^*\lambda)_x(u)=0$, this implies that $\lambda_x$ and $(\phi_t^*\lambda)_x$ have the same kernel and there exists a $f_t(x)$ such that $(\phi_t^*\lambda)_x=f_t(x)\lambda$. We deduce that ${d\over{dt}}_{t=0}={d\over {dt}}_{t=0}f_t\lambda=g\lambda$.
Conversely, if $L_X\lambda=g\lambda={d\over{dt}}_{t=0}\phi_t\lambda$, $\phi_t^*\lambda=\int_0^1g\lambda=f_t\lambda$ and $\phi_t^*\lambda$ and $\lambda$ have the same kernel.