(we denote by $X_1,\ldots,X_n$ indeterminates)
Let $A$ be subring of $R$ (both commutative). The elements $x_1,\ldots,x_n \in R$ are called algebraically independent over $A$, if for all Polynomials $\alpha \in A[X_1,\ldots,X_n]\setminus \{ 0 \}$ the equation \begin{align*} \alpha(x_1,\ldots,x_n) \neq 0 \end{align*} holds.
Question(A) Why is it important, that $A$ is a subring of $R$? Why don't we "simply" set $R=A[x_1,\ldots,x_n]$?
Lets define a homomorphism to get a claim about isomorphism: Say $a_1,\ldots,a_n \in R$ are algebraically independent over $A$ and $A$ is a subring of $R$. Then the map \begin{align*} \psi: A[X_1,\ldots,X_n] &\rightarrow R\\ \alpha &\mapsto \alpha(a_1,\ldots,a_n) \end{align*} is a homomorphism which is injective (obviously). The Image of $\psi$ is $A[a_1,\ldots,a_n]$. Thus $A[X_1,\ldots, X_n] \cong A[a_1,\ldots,a_n]$
Question (B) Why is the ring $R$ Important in this claim and why can't we leave it out?
The thing is that you have to get your $x_1,...,x_n$ from somewhere. They clearly can't lie in $A$ as then they would be algebraic over $A$, but you have to define a ringstructure on your elements, so the thing you do is that you take the $x_1,..x_n$ out of a ring extension $R$ of $A$, then $A$ is a subring of $R$.
This would be just the perspective concerning the definition. But in practice we are mostly interested in what we can say about ring extensions (or rings in relation to their subrings). This you can for example also see when looking at ringhomomorphisms $f: A \to R$, when $f$ is injective A can be seen as a subring of $R$, or if not $A/Ker(f)$, and then we can say things about that. So the situation where we look at a subring $A$ of $R$ and ask what elements are algebraicly independent over $A$ is something we encounter and are interested in.
One example of this would be Noether Normalization, if you are interested in that: https://en.wikipedia.org/wiki/Noether_normalization_lemma
Or you can just find abundant material in "Introduction to Commutative Algebra" by Atiyah and MacDonald.