In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)
We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $\overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $\mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $\mathbb Z$ of integers.
Let's examine the collapsing that takes place in the map $\pi: R \to \overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $\pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $\overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.
Any number of relations $a_1 = 0, \ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, \ldots, a_n$, the set of linear combinations $r_1 a_1 + \cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $\overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $\overline R$ if and only if $b'$ has the form $b + r_1 a_1 + \cdots +r_n a_n$ with $r_i$ in $R$.
The more relations we add, the more collapsing takes place in the map $\pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $\overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.
Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $\overline R = R / (a)$ be the result of killing $a$ in $R$. Let $\overline b$ be the residue of $b$ in $\overline R$. The Correspondence Theorem tells us that the principal ideal $(\overline b)$ of $\overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $\overline R / (\overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $\overline b$ in the ring $\overline R$ that is obtained by killing $a$ first.
Example 11.4.5. We ask to identify the quotient ring $\overline R = \mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $\mathbb Z[x] \to \mathbb Z[i]$ sending $x \mapsto i$ is the principal ideal of $\mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $\mathbb Z[x]/(f) \approx \mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $\overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $\mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $\overline R =\mathbb Z[x]/I$.
To form $\overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $\mathbb Z[x]$ is the kernel of the homomorphism $\mathbb Z[x] \to \mathbb Z$ that sends $x \mapsto 2$. So when we kill $x-2$ in $\mathbb Z[x]$, we obtain a ring isomorphic to $\mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $\overline R$ by killing $5$ in $\mathbb Z$, and therefore $\overline R \approx \mathbb F_5$.