I am trying to prove in a coordinate-independent way (see Fecko's "Differential Geometry and Lie Groups for Physicists, problem (18.3.2)) that
$$\theta_L := S(dL) = \hat{L}^* \theta \tag{1}\label{eq1}$$
where $\theta_L$ is the Cartan 1-form on $\pi:TM \to M$ associated to the Lagrangian function $L:TM \to \Bbb R$; $S :=\hat{1}^\uparrow$ is the vertical endomorphism (vertical lift of the ${1 \choose 1}$ unit tensor field on $M$), $\theta$ is the canonical 1-form on $\tau:T^*M \to M$ and $\hat{L}^* \theta$ is the pullback of $\theta$ induced by the the Legendre map $\hat L:TM \to T^*M$, defined by the relations ($v,w\in\pi^{-1}(x)$, $x \in M$): $$\tau \circ\hat L = \pi \qquad \langle \hat L(v),w \rangle :=w_v^\uparrow L \equiv \left.\frac{d}{dt} \right|_0 L(v+tw)$$ I understand that $$ \langle(\hat L^* \theta)_v,w \rangle= \langle \theta_{\hat L(v)}, \hat L_*w \rangle = \langle \hat L(v), \tau_* \hat L_*w \rangle= \langle \hat L(v), \pi_*w \rangle = \left( \pi_*w \right)^\uparrow L = \left.\frac{d}{dt} \right|_0 L(v+ t\pi_* w)$$
but cannot go on unless by using local canonical coordinates. Any hints?