How to prove $\{G_i\to F\}$ is open covering only if $\forall$ field $K$, $F(Spec K)=\cup_iG_i(Spec(K))$? This is an exercise in Eisenbud, Harris, Geometry of Schemes VI-11 as this part is skipped in Mumford Algebraic Geometry II. I think I figured out a way to do it but I am not totally sure. 
$\{G_i\to F\}$ is a collection of open subfunctors with $F:Schemes\to Set$ where open subfunctors means for all $h_R=Hom(-,Spec(R)),\phi\in Nat(h_R,F)$, $G_i\times_\phi h_R$ is a subfunctor of $h_R$ where pullback is defined on affine objects. 
Now $\{G_i\to F\}$ is called covering if for any scheme $X$ with $h_X=Hom(-,X)$ and any $\phi\in Nat(h_X,F)$, $G_i\times_Fh_X$ is representable as $h_{U_i}$ with $U_i$ covering $X$. 
Show that $\{G_i\to F\}$ is open covering iff $F(Spec(K))=\cup G_i(Spec(K))$ for all field $K$.
Forward direction is trivial by applying all functors to $Spec(K)$. The fiber product has either 1 element or none by embedding into $Hom(Spec(K), Spec(K))=Hom(K,K)=\{1_K\}$. It follows equality of $F(Spec(K))=\cup G_i(Spec(K))$. 
I am kind of having trouble with reverse direction. 
If $F$ is representable as a scheme, then it boils down to prove the statement for affine schemes where $G_i$ will be identified as hom functor of open subschemes of affine scheme. Use all residue fields to detect the missing points of covering. Then I can see it indeed forms a covering.
$\textbf{Q:}$ How do I prove the converse statement? I am also kind of having trobule to grasp the main point of the converse statement. What is the geometric meaning?
 A: It sounds like you've proved the hard part of the converse. I will take as proven that if $Y_i$ are open subschemes of a scheme $X,$ and for every field $K$ we have $X(K)=\bigcup Y_i(K),$ then $Y_i$ cover $X.$

Exercise VI-11. Let $\{G_i \to F\}$ be a collection of open subfunctors of a
  functor $F : \text{(schemes)} \to \text{(sets)}.$ Show that this is an open covering if and only if $F(\operatorname{Spec} K) =  \bigcup G_i(\operatorname{Spec} K)$ for all fields $K.$

Assume $F(\operatorname{Spec} K)=\bigcup G_i(\operatorname{Spec} K).$ Fix a scheme $X$ and $\phi:h_X\to F.$ There are open subschemes $Y_i$ of $X$ representing $h_X\times_F G_i.$ We need to show that these $Y_i$ cover $X.$
Consider a field $K.$ Every $K$-point $h_{\operatorname{Spec} K}\to h_X$ gives a $K$-point $h_{\operatorname{Spec} K}\to F$ by composing with $\phi.$ This factors through some $G_i\to F$ by hypothesis. By the property of a pullback, it therefore factors through $h_{Y_i}.$ We have shown that every $K$-point of $X$ is a $K$-point of one of the subschemes $Y_i,$ i.e. $X(K)=\bigcup Y_i(K).$ Since $K$ was an arbitrary field, this implies that $Y_i$ cover $X.$
This verifies that $\{G_i\to F\}$ is an open covering.
As for intuition: for the scheme case, you know that open covers can be detected by looking at $K$-points for all fields $K.$ Presheaves can be thought of as more general spaces, and it is similarly nice to know that open covers can still be detected by looking at their $K$-points.
