Localization as dual to quotienting?

The deeper / categorical lurking in the background is the notion of factorization structure. Let $E$ and $M$ be two classes of morphisms in a category. Then $(E,M)$ is said to be a factorization structure if:

1. $E$ and $M$ are closed under composition with isomorphisms;
2. every ring homomorphism has a factorization $\mu\circ\varepsilon$ with $\mu\in M$ and $\varepsilon\in E$;
3. the unique $(E,M)$-diagonalization property that for every commutative diagram$\require{AMScd}$ $$\begin{CD} A@>\varepsilon>>B\\ @V\varphi VV @VV\gamma V\\ C@>>\mu> D \end{CD}$$ with $\varepsilon\in E$ and $\mu\in M$ there exists a unique diagonal $\delta:B\to C$ making the diagram commtative.

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In the category of commutative rings we have the following factorization structures:

1. $E$ is the class of surjective ring homomorphisms and $M$ the class of injective ring homomorphisms;
2. $E$ be the class of ring localizations, that's of the form (up to isomorphism) $A\to S^{-1}A$ where $S\subseteq A$ is a multiplicative system of $A$ and $M$ be the class of ring homomorphism $\varrho:A\to B$ such that $A^\times=\varrho^{-1}(B^\times)$;
3. $E$ is the class of integral ring homomorphisms and $M$ is the class of injective and integrally closed ring homomorphisms.

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Consider the second factorization structure in the list above, every homomorphism of commutative rings $\varrho:A\to B$ has as essentially unique $(E,M)$ factorization $$A\xrightarrow\varepsilon S^{-1}A\xrightarrow\mu B$$ where $S=\psi^{-1}(R_2^\times)$. Thus, with your notation, we have to take $\operatorname{rek}(\psi)=\psi^{-1}(R_2^\times)=\{ r_1 \in R_1 : \psi(r_1) \in R_2^\times \}$.

Fabio gives a very nice answer to your question, but doesn't directly address an important point of confusion in your original post/comments, so I'm adding this answer for posterity. In general, a map $\psi:R_1\rightarrow R_2$ will absolutely not induce an epimorphism $\text{rek}(\psi)^{-1}R_1\twoheadrightarrow R_2$, even if we take the stronger definition $\text{rek}(\psi)=\psi^{-1}(R_2^\times)$ given by Fabio. For instance, if every element of $\text{rek}(\psi)$ is already a unit in $R_1$, then we will just have $\text{rek}(\psi)^{-1}R_1=R_1$, and so using this it is easy to come up with examples where the induced map is not epi.

For instance, take $R_1=\mathbb{Q}$, and $R_2$ any field extension of $\mathbb{Q}$ with a non-trivial automorphism $\alpha$ that fixes $\mathbb{Q}$ pointwise, with $\psi:R_1\hookrightarrow R_2$ the inclusion map. Then $\text{rek}(\psi)=\mathbb{Q}^\times$ , so $\text{rek}(\psi)^{-1}\mathbb{Q}=\mathbb{Q}$ and the induced map to $R_2$ is just $\psi$, which is certainly not an epimorphism. (E.g. $\alpha\circ\psi=\text{id}_{R_2}\circ\psi$ but $\alpha\neq\text{id}_{R_2}$).

Indeed, the polynomial ring example you give in the comments of your post does not hold in general either. If we let $R_1=\mathbb{R}[x]$ and $R_2=\mathbb{R}[x,y]$, with $\psi:R_1\hookrightarrow R_2$ again the inclusion map, then once again $\text{rek}(\psi)=R_1^\times$ but $\psi$ is certainly not epi.

The problem in all of these examples is that $R_2$ can be very big compared to the image of $R_1$; hopefully the above examples clarify that point. (Note however, that – provided $R_2\neq\{0\}$ – the map $R_1\hookrightarrow \text{rek}(\psi)^{-1}R_1$ will still be injective, even if we use Fabio's stronger definition of $\text{rek}(\psi)$, because no element of $\text{rek}(\psi)$ can be a zero-divisor in $R_1$.)