What are the quadratic extensions of $\mathbb{Q}_2$? How do you classify the non-squares in $\mathbb{Q}_2$? I've tried writing down expansions for "odd" numbers in $\mathbb{Z}_2$, but unlike in $\mathbb{Z}_p$, the n$^{th}$ term in the expansion is not always uniquely determined once you know the first n-1 terms, and when you come across a non-square it's not obvious (to me at least) whether it is divisible by another non-square you've already found.

As for any field $K$ of characteristic different from $2$, the quadratic extensions are all of the form $K(\sqrt{d})$ for $d \in K^{\times} \setminus K^{\times 2}$; moreover $K(\sqrt{d_1}) \cong K(\sqrt{d_2}) \iff d_1 = a^2 d_2$.  Thus they are parameterized by the nontrivial elements of $K^{\times}/K^{\times 2}$, it is then sufficient to understand this quotient.  Note that this is an $\mathbb{F}_2$-vector space, so it's enough to determine its dimension.  In what follows I will denote this $\mathbb{F}_2$-dimension simply by "$\operatorname{dim}$".
If $K$ is the fraction field of a discrete valuation ring $R$, then from $K^{\times} \cong R^{\times} \oplus \mathbb{Z}$ it is easy to see that
$\dim K^{\times}/K^{\times 2} =  1+ \dim R^{\times}/R^{\times 2}$.
So, here, you want to know the square classes in $\mathbb{Z}_2^{\times}$.  I claim that an element $u \in \mathbb{Z}_2^{\times}$ is a square iff its residue modulo $8$ is a square in $\mathbb{Z}/8\mathbb{Z}$: to see this, use Hensel's Lemma.  From this it follows that 
$\dim \mathbb{Z}_2^{\times} / \mathbb{Z}_2^{\times 2} = \dim (\mathbb{Z}/8\mathbb{Z})^{\times} / (\mathbb{Z}/8\mathbb{Z})^{\times 2} = 2$
and thus
$\dim \mathbb{Q}_2^{\times} / \mathbb{Q}_2^{\times 2} = 3$.
This argument should give you explicit representatives as well: that is, the $2^3-1 = 7$ quadratic extensions of $\mathbb{Q}_2$ are gotten by adjoining square roots of $3,5,7,2,6,10,14$.  
