Sheafification is left adjoint to forgetful functor

This is a sillier question but I'm working through Vakil's notes, exercise 2.4L and I'm having trouble seeing why sheafification is left adjoint to the forgetful functor.

I have done the exercises up till now and see why sheafification as constructed satisfies the universal property. But I don't see how the universal property gives us the adjointness a-priori.

As far as I understand, for the two functors to be an adjoint pair, we need to have a bijection

$$\phi : Mor(\mathcal{F}^{sh}, \mathcal{G}) \rightarrow Mor(\mathcal{F}, \mathcal{G}^{pre})$$.

Where $$\mathcal{G}$$ is a sheaf and $$\mathcal{F}$$ is a presheaf. Now I know that the universal property allows us to construct a map of sheaves $$\mathcal{F}^{sh} \rightarrow \mathcal{G}$$ from a map of presheaves $$\mathcal{F}^{pre} \rightarrow \mathcal{G}$$. So this is the backwards map in the bijection. But what is the forward map? Is every map from the sheafification of $$\mathcal{F}$$ to $$\mathcal{G}$$ an induced map of the presheaves obtained from the universal property? Surely we don't just apply the forgetful functor to $$\mathcal{F}^{sh}$$ and $$\mathcal{G}$$ and take the same map, since sheafification and forgetful functor aren't inverses.

What's the exact map from the left side to the right side?

$$\DeclareMathOperator{\sh}{sh}\DeclareMathOperator{\pre}{pre}$$Fix a presheaf $$\cal F$$, and a sheaf $$\cal G$$. The mantra about sheafification is: "a morphism of sheaves $$\cal F^{\sh} \to \cal G$$ is the same as a morphism of presheaves $$\cal F \to \cal G$$".
(In your notation, the latter is written as $$\cal F \to \cal G^{\pre}$$.)
From what you've written, you seem to understand how a presheaf morphism $$\cal F \to \cal G$$ gives a sheaf morphism $$\cal F^{\sh} \to \cal G$$.
Indeed, recall that the sheafification is not just a sheaf $$\cal F^{\sh}$$ but also a presheaf morphism $$\cal F \xrightarrow{\sh} \cal F^{\sh}$$. Thus, given a morphism of sheaves $$\cal F^{\sh} \to \cal G$$, it is also a morphism of presheaves and composing as $$\cal F \xrightarrow{\sh} \cal F^{\sh} \to \cal G$$ gives us the presheaf morphism that we wanted.