Euler-Lagrange's equations for a Lagrangian $L$ read $$\frac{d}{dt}\frac{\partial L}{\partial \bf v} - \frac{\partial L}{\partial \bf x} = 0.$$ More precisely, the statement is that a trajectory $\gamma:\mathbb R\to M$ that minimises the corresponding action $\gamma\mapsto \int L\circ d\gamma$, satisfies the equation $$A_L[\gamma] \equiv (\partial_{\bf v}L\circ d\gamma)' - (\partial_{\bf x}L)\circ d\gamma = 0.\tag1$$ Here, $M$ is the configuration space; I'm thinking of the Lagrangian as a map $L:TM\to\mathbb R$ defined on the tangent bundle of $M$ (though I suppose this is strictly the case only when $L$ does not depend explicitly on the time), and the differential is understood as the mapping $$d\gamma:\mathbb R\to TM:t\mapsto (\gamma(t),d\gamma_t(\partial_t))\simeq (\gamma(t),\gamma'(t)).$$

In (1), I defined the operator $A_L$ mapping curves $\gamma:[a,b]\to M$ into functions $[a,b]\to\mathbb R$. Euler-Lagrange's equations, in terms of this operator, thus say that if $\gamma$ is stationary for the action, then $A_L[\gamma]$ is the zero function.

Is there a way to understand this operator $A_L$ on more general/abstract/geometric grounds? More specifically, thus the specific structure of the differentials in the expression $\frac{d}{dt}\frac{\partial}{\partial \bf v}-\frac{\partial}{\partial\bf x}$ somehow arise geometrically?