Let $(x_0,x'_0) \in \mathbb{R}^2$ be initial data for the Euler Lagrange equation with some given Lagrangian $L: \mathbb{R}^2 \rightarrow \mathbb{R}.$
Then $F^t$ is the flow that maps the initial data to the solution of the Euler Lagrange equation for a given time $t$.
The action is given by $$S_t(x_0,x_0') = \int_0^t L(F^{\tau}(x_0,x_0')) d\tau$$
Calculating the directional derivative of this function $S_t$ at some $(x_0,x_0')$ in direction $(\xi,\xi')$ gives us then
$$dS_t(x_0,x_0')(\xi,\xi') = \partial_2L(F^t(x_0,x_0')) \xi(\tau)|_{\tau=0}^{\tau=t}.$$
$\xi(\tau)$ is here a short notation for the propagation of the Euler Lagrange equation with initial data $\xi.$
So far, this is alright. Now, there is a step where I cannot follow:
Let $\theta = \partial_2L(x_0,x_0') dx$ be a 1-form, then it is claimed that
$dS_t = (F^{t})^{*}(\theta) - \theta.$ I see that the term $\theta(x_0,x'_0)(\xi,\xi') = \partial_2L(F^0(x_0,x_0')) \xi$ and therefore $\theta = \partial_2L(x_0,x_0')dx$, so the last term is alright, but what about the first one $(F^t)^*(\theta)$?