There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.
First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-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 $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$.
Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. 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, for each $t\in[a,b]$ consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.
This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.