Let $X$ be a configuration space and $L: TX \rightarrow \mathbb{R}$ a Lagrangian. Then I want to show that the Lagrangian flow $F^t(x(0),x'(0)) = (x(t),x'(t))$ preserves the symplectic form just like the Hamiltonian flow does.


My idea was to use the Euler Lagrange equations.


So let $F^t(x(0) + \varepsilon \xi(0),x'(0)+ \varepsilon \xi'(0)) = (x\_{\varepsilon}(t),x'\_{\varepsilon}(t))$ be the flow of the Euler Lagrange equations.


Then we have after the usual integration by parts trick that
$$dS\_t(x(0),x'(0))(\xi(0),\xi'(0)) := d\_{\varepsilon}|\_{\varepsilon=0} \int\_0^t L(x\_{\varepsilon}(\tau),x'\_{\varepsilon}(\tau)) d\tau = \sum\_{k=1}^{n} \partial\_2L(F^\tau((x(0),x'(0))) \xi\_k(\tau)|\_0^t.$$


Note, that here the Euler Lagrange equation vanishes due to the property of the flow being a solution to the equations for all times. 


This last equation almost looks like $dS\_t = (F^t)^\* \theta - \theta,$ 
where $\theta = \sum\_{k=1}^{n} (\partial\_2 L)\_k dx^k$ which would give us of course the result, because then $0=d(dS\_t) = (F^t)^\* \omega - \omega$ by the usual commutation of the pullback with the exterior derivative.


For this we would need $(F^t)^\*\omega(x(0),x'(0))(\xi(0),\xi'(0)) = \sum\_{k=1}^{n} \partial\_{x\_k'(0)} L(F^t(x(0),x'(0)))\xi\_k(t)$ holds which is only the case if this $\xi(t)= dF^t\_{x,x'}(\xi(0),\xi'(0))$ according to my notation above, but I don't see that this equality holds.



Lagrangian flow preserves symplectic form
You should indeed come to the conclusion that $dS\_t=F\_t^\*\theta-\theta$. 


Note that your $1$-form $\theta$ can be written in terms of the summation convention as $\theta=\frac{\partial L}{\partial \dot{x}^k}\,dx^k$. Now fix $(x,\dot{x})\in TX$ and let $(x(t),\dot{x}(t))=F^t(x,\dot{x})$. Also fix a tangent vector $(\delta x,\delta\dot{x})\in T\_{(x,\dot{x})}X$ and set 
$$(\delta x(t),\delta\dot{x}(t))=T\_{(x,\dot{x})}F^t(\delta x,\delta\dot{x}).$$
The pullback of $\theta$ along $F^t$ can be written


$$[(F^t)^\*\theta](\delta x,\delta\dot{x})=\frac{\partial L}{\partial\dot{x}^k}(x(t),\dot{x}(t))\delta x^k(t).$$


Next notice that your function $S\_t:TX\rightarrow\mathbb{R}$ can be written


$$S\_t(x,\dot{x})=\int\_0^tL(x(\tau),\dot{x}(\tau))\,d\tau.$$
As you noticed, only the boundary terms contribute to $dS\_t$, in particular,
$$dS\_t(\delta x,\delta\dot{x})=\frac{\partial L}{\partial \dot{x}^k}(x(t),\dot{x}(t))\delta x^k(t)-\frac{\partial L}{\partial x^k}(x,\dot{x})\delta x^k=(F^{t\*}\theta-\theta)(\delta x,\delta\dot{x}).$$ 


