In page 13 of this course https://www.sissa.it/fa/download/publications/remizov.pdf from the equation $$ (F_x+pF_y) dx + F_pdp=0 \qquad (*) $$ I do not understand how we can get $$ \dot x =F_p, \quad \dot p = -(F_x+pF_y) \qquad (**). $$
($F_j$ denotes the derivative wrt $j\in \{x,y,p\}$)
As far as I understand, $(*)$ is an equation where $dx,dy$ are linear form (differential 1-forms), while $(**)$ defines vector fields.
In (*) you are looking for vector fields $X=(u,v)$ with $$ Mu+Nv=0, $$ using $dx(X)=u$, $dp(X)=v$. This can be interpreted as an orthogonality condition. Thus one trivial solution is to rotate $(M,N)$ by 90° to get $(u,v)=(N,-M)$ or $(u,v)=(-N,M)$. You can also scale by any other non-zero factor (function).
The resulting ODE system, with $\dot y=p\dot x$, gives solutions $x(t),y(t),p(t)$. On segments where $x(t)$ is monotonic it can be inverted, and then also $y$ and $p$ can be obtained as functions of $x$.