Lagrangian vector field expression The Lagrangian vector field $X_L$ on the tangent bundle is given in page 4 of Marco Mazzucchelli's "critical Point Theory for Lagrangian systems" to be;
\begin{equation}
X_L=\sum^M_{j=1}\bigg(v^j\frac{\partial}{\partial q^j}+\sum ^m_{h=1}\bigg[\frac{\partial ^2L}{\partial v^j\partial v^h}\bigg]^{-1}\bigg(\frac{\partial L}{\partial q^h}-\frac{\partial ^2L}{\partial t\partial v^h}-\sum^m_{l=1}\frac{\partial^2 L}{\partial q^l\partial v^h}v^i\bigg)\frac{\partial }{\partial v^j}\bigg).
\end{equation}
My question is ... how does one arrive at this expression? I have started by trying to compute an expression from the following;
\begin{equation}
X_{\mathscr L}=\sum _{i=1}^n \dot q^i\frac{\partial }{\partial q^i}+\sum ^n_{i=1}\ddot q^i\frac{\partial }{\partial \dot q^i}.\nonumber
\end{equation}
With no luck! 
 A: I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums.
Let
\begin{equation}
 \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} 
\end{equation}
be a lagrangian over $T M$, with time in $\mathbb{R}$. Euler-Lagrange equations say that
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q}(q,\dot{q};t) - \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}}(q,\dot{q};t) = 0 .
\end{equation}
Take now the total derivative and expand it:
\begin{equation}
 \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}}(q,\dot{q};t) = \frac{\partial^2 \mathcal{L}}{\partial \dot{q} \partial \dot{q}} \frac{d \dot{q}}{dt} + \frac{\partial^2 \mathcal{L}}{\partial q \partial \dot{q}} \frac{d q}{dt} + \frac{\partial^2 \mathcal{L}}{\partial t \partial \dot{q}} \frac{d t}{dt} .
\end{equation}
Substituting that in E-L equations and reorganising gives us
\begin{equation}
\frac{\partial^2 \mathcal{L}}{\partial \dot{q} \partial \dot{q}} \frac{d \dot{q}}{dt}= \frac{\partial \mathcal{L}}{\partial q} - \frac{\partial^2 \mathcal{L}}{\partial q \partial \dot{q}} \frac{d q}{dt} - \frac{\partial^2 \mathcal{L}}{\partial t \partial \dot{q}} .
\end{equation}
Now me must assume the second derivative of $\mathcal{L}$ with respect to the velocity is different from zero. This, on higher dimensions, means that the jacobian of the derivative mapping of $\mathcal{L}$ must be nonsingular. If that is true we can say that
\begin{equation}
\ddot{q} = \left[\frac{\partial^2 \mathcal{L}}{\partial \dot{q} \partial \dot{q}}\right]^{-1}\left( \frac{\partial \mathcal{L}}{\partial q} - \frac{\partial^2 \mathcal{L}}{\partial q \partial \dot{q}} \frac{d q}{dt} - \frac{\partial^2 \mathcal{L}}{\partial t \partial \dot{q}} \right) .
\end{equation}
Now, as you've said, the field generated by $\mathcal{L}$ is
\begin{equation}
 X_\mathcal{L} = \dot{q} \frac{d}{dq} + \ddot{q} \frac{d}{d \dot{q}} ,
\end{equation}
and substituting what we've found we get to
\begin{equation}
X_\mathcal{L} = \dot{q} \frac{d}{dq} + \left[\frac{\partial^2 \mathcal{L}}{\partial \dot{q} \partial \dot{q}}\right]^{-1}\left( \frac{\partial \mathcal{L}}{\partial q} - \frac{\partial^2 \mathcal{L}}{\partial q \partial \dot{q}} \frac{d q}{dt} - \frac{\partial^2 \mathcal{L}}{\partial t \partial \dot{q}} \right) \frac{d}{d \dot{q}} ,
\end{equation}
which is a 1-D version of what you've posted.
