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?

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.