The first Thing I have done was to verify that $D$ is a normal subgroup. I did so by verifying that
$$f:G\times G \rightarrow G, f(a,b)=a^{-1}b$$
is linear, and that the kernel of $f$ is $D$.
How can I answer now the Question in the title?
Because I only know that $G/D\cong \text{Im}(f)$
which is the sign for bijectivity but not for equality.
Let's take a concrete example -- where $G$ is the integers. So $H = G \times G$ consists of all pairs of integers. The relation defined by $D$ says that two items, $(a, b)$ and $(c, d)$ in $H$ are $D$-equivalent if there's some element $(g,g) \in D$ with $$ (a+g, b+g) = (c, d) $$
Alternatively, we can look at one element $(a, b) \in H$ and ask "What are all the things in its equivalence class (mod $D$)?"
We add each item in $D$ to $(a,b)$ to get $$ [(a,b)] = \{ (a+n, b+n) \mid n \in \Bbb Z \} $$ In other words, thinking geometrically, where $\Bbb Z \times \Bbb Z$ is the integer grid in the plane, an equivalence class is exactly a 45-degree diagonal in the integer grid. The equivalence class for $(0,0)$ is exactly the points $$ \ldots, (-2, -2), (-1, -1), (0, 0) , (1, 1) , (2, 2), \ldots, $$ for instance. And any other equivalence class is just this diagonal set, moved up or down.
One thing that means is that every equivalence class contains exactly one element on the $x$-axis...in particular, $[(a, b)]$ contains $(a-b, 0)$. The map taking $H/D$ to $G$ by sending $[(a, b)]$ to $a-b$ is therefore well-defined, and it turns out to be an isomorphism.
So:
i. Elements of $G/D$ look like diagonals in $G \times G$.
ii. The group $G/D$ is isomorphic to $G$ via $(a, b) \mapsto a-b$ (or any of many other isomorphisms.