Gaussian Integers form an Euclidean Ring

You seem to be a bit confused. $b$ could be any element of $G$ except $0$. You have already found a good choice for $f$ so the only part left is, given arbitrary $a, b, \in G$ with $b \neq 0$, to find $q, r \in G$ satisfying (i) and (ii).

It might help you to understand what can happen when a ring is not Euclidean. For example, consider $\mathbb Z[\sqrt{-5}]$, consisting of all numbers of the form $m + n \sqrt{-5}$, with $m, n \in \mathbb Z$.

The natural choice of Euclidean function is $$f(m + n \sqrt{-5}) = m^2 + 5n^2.$$ Note that $f(m + n \sqrt{-5})$ can never be negative, but it can be 0. Now try $\gcd(2, 1 + \sqrt{-5})$.

Since $f(1 + \sqrt{-5}) > f(2)$, we assign $a = 1 + \sqrt{-5}$ and $b = 2$. Now we wish to express $a = qb + r$ with $f(r) < f(b)$, meaning $f(r) < 4$. Then the only possibilities for $r$ are $-1$, 0 and 1, as all other numbers in this ring have norms of at least 4.

But then we see that none of the numbers $\sqrt{-5}$, $1 + \sqrt{-5}$, $2 + \sqrt{-5}$ are divisible by 2. Therefore the Euclidean algorithm has failed in this ring, and therefore this domain is not Euclidean for the function $f$.

Your task then is to prove that this can never happen in $G$, or $\mathbb G$, or $\mathbb Z[\sqrt{-1}]$, or as it's more commonly notated, $\mathbb Z[i]$, to prove that a suitable choice of $r$ can always be found. The Euclidean function is $f(m + ni) = m^2 + 1n^2$.

A complete solution is to be found in certain number theory books, such as An Introduction to the Theory of Numbers by Ivan Niven and Herb Zuckermann.