Given $F(x,y)$, what does it mean to compute $dF(X)$ for $X=x \frac{d}{dx}+y\frac{d}{dy}$? Given $F(x,y)$, what does it mean to compute $dF(X)$ for $X=x \frac{d}{dx}+y\frac{d}{dy}$?
My idea:
$dF(x,y)$ is the same as the Jacobian of $F(x,y)$. But in order to plug in $X$, then what should I do?

By definition the differential $\mathrm{d}F_p$ of the function $F$ at a point $p$ is a linear operator taking in input a tangent vector $X_p$ and returning the directional derivative of $F$ along $X_p$ at the point $p$, i.e. 
$$\mathrm{d}F_p(X_p):=X_pF = \frac{\mathrm{d}F}{\mathrm{d}t}(\gamma(t))\Big\rvert_{t=0},$$
where $\gamma\colon (-\varepsilon,\varepsilon) \to \mathbb{R}^2$ is a smooth curve such that $\gamma(0)=p$ and $\gamma'(0)=X_p$. 
From this definition you can see, for example, that $$\mathrm{d}x\left(\frac{\partial}{\partial x}\right) = \frac{\partial x}{\partial x}=1, \quad \mathrm{d}x\left(\frac{\partial}{\partial y}\right)=\frac{\partial x}{\partial y} =0.$$
Therefore, if $X=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$ and $p=(x,y)$ you get
$$dF_p(X_p)= X_pF=x\frac{\partial F}{\partial x}(x,y) +y\frac{\partial F}{\partial y}(x,y).$$
