$R$ is the set of all real number.
$R[x , y] \over \langle x^2 +1\rangle$ isomorphic to $R[i] [y]$
$R[x , y] \over \langle x\rangle $ isomorphic to $R[0][y]$
then why $R[x , y] \over \langle x^2\rangle$ is not isomorphic to $R[0][y]$? Rather it is isomorphic to $R[i][y]$. Despite the fact that $0$ is a root of $x^2$.
Can anyone explain me in simple language keeping in mind the fact that I only have the basic knowledge of Ring ?
Thank You in advance.
First of all, everywhere where you wrote $R[x,y]$, you should instead have $R[x]$.
You seem to be working under the misunderstanding that $R[x]/\langle p(x)\rangle$ is equal to $R[\alpha]$, where $\alpha$ is a root of $p(x)$. This is only true when $p(x)$ is an irreducible polynomial over $R$, meaning it cannot be factored into smaller real polynomials. The polynomial $x^2+1$ is irreducible over $R$, so you can say $R[x]/\langle x^2+1 \rangle=R[i],$ where $i^2+1=0$. Note that $x^2+1$ is reducible over $\mathbb C$, but that does not matter.
In your case, $x^2=x\cdot x$ is reducible, so you cannot say that $R[x]/\langle x^2\rangle=R[0]$. The best you can say is that every element of $R[x]/\langle x^2\rangle$ is of the form $ax+b$, where $x$ is a symbol which satisfies $x\neq 0$ and $x^2=0$.