How are the following two rings isomorphic? The paper here in Construction 4.16 makes the following claim that I'm unable to unpack (though my question should be self-contained here):
Let $R$ be a commutative ring, and let $r\in R$ be an element that is not a zero divisor. Let $S=R/(r)$. There is evidently an isomorphism between $S$ and the tensor product $R\otimes_{R[y]}R$, where the two factors are $R[y]$-algebras in two different ways. Those two different ways are given by the pushout square
$
\require{AMScd}
\begin{CD}
R[y] @>{\lambda}>> R;\\
@V{\psi}VV @VVV \\
R @>{\varphi}>> R\otimes_{R[y]}R;
\end{CD}
$
The map $\lambda:R[y]\to R$ is given by $y\mapsto0$ and the map $\varphi$ is defined in the paper as the canonical surjection $\varphi:R\to S$. The issue is that I don't know what the map $\psi:R[y]\to R$ is, nor the unlabeled map $R\to R\otimes_{R[y]}R$, nor how to think of $\varphi$ if its target is $R\otimes_{R[y]}R$.
As a note, the tensor product is written in the paper as $R^{\psi}\otimes_{R[y]}{}^{\epsilon}R$, where $\epsilon$ is never defined. (I think it is a typo that should be $\lambda$.) If I understand the notation correctly, the superscript maps are indicating in what way $R$ is an $R[y]$-algebra (i.e., $\psi:R[y]\to R$ and $\epsilon/\lambda:R[y]\to R$).
So my questions are:

What is the map $\psi:R[y]\to R$? In some way, this should be where we realize quotienting by $r$.
What is the unlabeled map $R\to R\otimes_{R[y]}R$? As a pushout, it should be $x\mapsto1\otimes x$?
What is the map $\varphi:R\to R\otimes_{R[y]}R$? As a pushout, it should be $x\mapsto x\otimes1$?
And what is the isomorphism $S\cong R\otimes_{R[y]}R$?

Thank you!
 A: 
What is the map $\psi:R[y]\to R$? In some way, this should be where we realize quotienting by $r$.

It is the evaluation map that sends $y$ to $r$.

What is the unlabeled map $R\to R\otimes_{R[y]}R$? As a pushout, it should be $x\mapsto1\otimes x$?
What is the map $\varphi:R\to R\otimes_{R[y]}R$? As a pushout, it should be $x\mapsto x\otimes1$?

That is correct.

And what is the isomorphism $S\cong R\otimes_{R[y]}R$?

To avoid confusion of notation, let me write $A$ for $R$ as an $R[y]$-algebra via $\psi$ and $B$ for $R$ as an $R[y]$-algebra via $\lambda$.  Note that there are $R[y]$-algebra isomorphisms $A\cong R[y]/I$ and $B\cong R[y]/J$ where $I=\ker\psi=(y-r)$ and $J=\ker\lambda=(y)$.  So, $A\otimes_{R[y]} B\cong R[y]/(I+J)$.  But $I+J$ is just the ideal $(y,r)$, so $R[y]/(I+J)=R[y]/(y,r)\cong R/(r)=S$.

Or, in terms of universal properties: a pair of homomorphisms $f,g:R\to T$ such that $f\psi=g\lambda$ must satisfy $f=g$ (since $\psi$ and $\lambda$ are the identity on constants in $R[y]$) and also $f(\psi(y))=f(r)$ must be the same as $g(\lambda(y))=g(0)=0$, and conversely $f=g$ and $f(r)=0$ are enough to guarantee that $f\psi=g\lambda$.  So, homomorphisms out of the pushout are in natural bijection with homomorphisms out of $R$ that send $r$ to $0$, and those are in natural bijection with homomorphisms out of $R/(r)$.
