From here:
Why is in the sub-example with $Y\cap Z$, $(Y\sqcup Z)/{\sim}=Y\cup Z$? The text uses the undefined terminology ("to glue" a subset to a subset).
Also, the text claims that I an check that the square depicted is a pushout "by verifying the universal property or by using the formula just stated". What is "the formula just stated"? I don't see any formulas. As for verifying the universal property, as far as I understand, it will follow automatically from this after one verifies that $(Y+Z)/{\sim}=Y\cup Z$, right?
The 'formula just stated' refers to example 5.2.12, where it gives the general formula of a pushout in $\bf Set$.
Here we have $s:Y\cap Z\to Y$ and $t:Y\cap Z\to Z$ the inclusions, and thus the equivalence relation on $Y\sqcup Z$ is generated by $s(x) \sim t(x)$ for $x\in Y\cap Z$, that is, the $Y$-copy of $x$ is equivalent to ('glued to') the $Z$-copy of the same $x\in Y\cap Z$.
No other elements of $Y\sqcup Z$ are affected by this relation, so that their equivalence classes are singletons.
The other suggested method is to verify the universal property right away for the union $Y\cup Z$.
For that, consider two maps $Y\to W$ and $Z\to W$ which agree on $Y\cap Z$, and show they induce a unique map $Y\cup Z\to W$ (which makes the appropriate diagrams commute).
Combined with the first method, it also proves $Y\cup Z\ \cong \ Y\sqcup Z/\sim$.