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I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer).

Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space , $T^*Q$, and doesn't depend on the Hamiltonian $H$.

Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be `hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$.

I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his).

Edit: I should clarify that by `Legendre transform' above I mean (in coordinates) the map $p_i(x, \dot{x}) = \frac{\partial L}{\partial \dot{x}^i}(x, \dot{x})$. This is standard in the literature, but it differs from the classical meaning $H(x, p) = p\dot{x}-L(x, \dot{x})$ (where $p_i = \frac{\partial L}{\partial \dot{x}^i}$).
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I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer).

Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space, and doesn't depend on the Hamiltonian $H$.

Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be `hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$.

I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his).

Edit: I should clarify that by `Legendre transform' above I mean (in coordinates) the map $p_i(x, \dot{x}) = \frac{\partial L}{\partial \dot{x}^i}(x, \dot{x})$. This is standard in the literature, but it differs from the classical meaning $H(x, p) = p\dot{x}-L(x, \dot{x})$ (where $p_i = \frac{\partial L}{\partial \dot{x}^i}$).
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I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer).

Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space, and doesn't depend on the Hamiltonian $H$.

Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be `hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$.

I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his).