If we have a Hamiltonian $$ H(x_1,\ldots,x_n;p_1,\ldots,p_n) $$ and Hamilton's equations $$ \frac{d x_i}{dt}=\frac{\partial H}{\partial p_i},\quad\frac{dp_i}{dt}=-\frac{\partial H}{\partial x_i}\tag{1} $$ it is said that thus the smooth function $H$ gives us a vector field $$ \xi_H=\sum \frac{\partial H}{\partial p_i}\frac{\partial}{\partial x_i}-\sum\frac{\partial H}{\partial x_i}\frac{\partial}{\partial p_i}\tag{2} $$
I do not see the connection between (1) and (2). How do equations (1) give vector field (2)?
As a hint:
You should try to rewrite the flow equation of this vector field, that is $$\frac{\text{d}u}{\text{d}t}=\xi_H\circ u$$ for a trajectory $u$, in terms of the local coordinates $x_1,\dots,x_n,p_1,\dots,p_n$. By some condition (which you can look up on wiki) this is the unique vector field providing your equations.
Edit:
Suppose your Hamiltonian is a function $H:\mathbb{R}^{2n}\to\mathbb{R}$. Then the trajectory of the system governed by this Hamiltonian is a map $\mathbb{R}\to\mathbb{R}^{2n}$, write its components as $$t\mapsto u(t)=\big(x_1(t),\dots,x_n(t),p_1(t),\dots,p_n(t)\big).$$ The derivative of this trajectory is now a vector, to be computed by $$\frac{\text{d}u}{\text{d}t}=\frac{\text{d}x_1}{\text{d}t}\frac{\partial}{\partial x_1}+\dots+\frac{\text{d}x_n}{\text{d}t}\frac{\partial}{\partial x_n}+\frac{\text{d}p_1}{\text{d}t}\frac{\partial}{\partial p_1}+\dots+\frac{\text{d}p_n}{\text{d}t}\frac{\partial}{\partial p_n},$$ where the $\frac{\partial}{\partial x_i},\frac{\partial}{\partial p_j}$ are a basis of (tangent) vectors. The vector field $\xi_H$ you have written is given in the same basis, so comparing basis coefficient gives you get back the Hamilton equation. The other way round, inserting the Hamilton equations in the above representation of $\frac{\text{d}u}{\text{d}t}$ defines your $\xi_H$. All this works the very same if you replace $\mathbb{R}^{2n}$ by the cotangent bundle $T^*M$ of some manifold with the correct coordinates, and some formulas may be rewritten more geometric (cf. the other answer).
This should answer your original question, I guess, but I don't understand your comment below. Should $F$ be an observable?