I think "universal example" here is an abuse of language to mean "example of a universal property".
As pointed out by t.b. in the comments:
By definition, an $R$-algebra for a commutative ring $R$ is a pair $(S, f)$ where $f: R \to S$ is a ring homomorphism and $S$ is a commutative ring.
An $R$-algebra $A$ with $n$ distinguished elements is hence a triplet $((a_1, \dots, a_n), A, f)$ for some $a_i \in A$ where $f: R \to A$ is a ring homomorphism. An $R$-algebra homomorphism between two $R$-algebras with $n$ distinguished elements $((a_1, \dots, a_n), A, f)$ , $((b_1, \dots, b_n), B, f^\prime)$ is an $R$-algebra homomorphism $\varphi : A \to B$ such that $\varphi (a_i) = b_i$.
Then the universal property that's stated is:
An $R$-algebra with $n$ distinguished elements is a triplet $((X_1, \dots, X_n), P, \varphi)$ satisfying the following universal property: for every $R$-algebra $((r_1^\prime, \dots, r_n^\prime), R^\prime, \phi)$ with $n$ distinguished elements $r_1^\prime, \dots, r_n^\prime \in R^\prime$ there exists a unique $R$-algebra homomorphism $\pi: P \to R^\prime$ such that $\pi\restriction_R = \pi \circ \varphi = \phi$ and $\pi (X_i) = r_i^\prime$ for the $n$ distinguished elements $r_i^\prime \in R^\prime$ and $n$ distinguished elements in $P$. Can't resist to add the corresponding diagram:
In general, many universal mapping properties read as follows:
The thingamajig is an object $B$ and a morphism $m: A \to B$ such that for every object $C$ and morphism $n: A \to C$ there is a unique morphism $\varphi : B \to C$ such that the diagram of morphisms commutes, that is, such that $\varphi \circ m = n$.