There is a coordinate-free description using just natural objects on $TM$.  Here is one way to do it.

First, consider the base point submersion $\pi:TM\to M$.  For each $v\in TM$, the map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is a submersion, and the fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space.  It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$.  Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)$. 

Now consider a Lagrangian $L:TM\to \mathbb{R}$. Define a $1$-form $\omega_L$ on $TM$ by 
$$
\omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr)
$$ 
for all $w\in T_v(TM)$ and $v\in TM$.  Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$.  (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then consider the $1$-form define by the rule
$$
\beta(w) = d\omega_L(\gamma''(t),w)+ dE_L(w)
$$ 
for $w\in T_{\gamma'(t)}TM$.  It is not hard to show that $\beta(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta = (\pi'(\gamma'(t))^\ast(\delta(\gamma(t)))$ for a unique $\delta(\gamma(t))\in T^\ast_{\gamma(t)}M$.  

This assignment $t\mapsto \delta(\gamma(t))$ is the canonical `variation' $1$-form.  It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.