Let $W(k)$ be the Witt-Ring of the field $k$.
in this script http://alpha.math.uga.edu/~pete/quadraticforms2.pdf at the bottom of page 2 the signed determinant is introduced by
$d^\pm (q) = (-1)^{n(n-1)/2}d(q) \in k^*/k^{*2}$, while $q \in W(k)$ and $n = rank(q)$.
This is done in order define the discriminant on $W(k)$ such that it is well defined.
Lets take $k = \mathbb{Q}_p$. There are two cases.
1 . $\sqrt{-1} \notin \mathbb{Q}_p$
In this case we have $-1 \in \mathbb{Q}^*_p/\mathbb{Q}^{*2}_p $ and $d(\mathbb{H}) = -1$ is well defined.
2 . $\sqrt{-1} \in \mathbb{Q}_p$
Now $-1$ is square an therefore $-1 \notin \mathbb{Q}^*_p/\mathbb{Q}^{*2}_p$. But $\mathbb{H} = X^2 - Y^2 \simeq X^2 + Y^2$ and so $d([\mathbb{H}]) = 1$ is defined and we have trouble by just looking at the isometric classes of forms. I dont get why there has to be a signed determinant at this point. This usual discriminant IS well defined on $W(k)$ and up to square classes. Or is it unnecessary in the case of $k =\mathbb{Q}_p$ ?
You seem a little confused about the difference between the Grothendieck-Witt ring $\hat{W}(K)$ and the Witt ring $W(K)$. Quadratic forms live in $\hat{W}(k)$; two forms $q_1,q_2 \in \hat{W}(K)$ become equivalent in $W(K)$ iff there are natural numbers $a_1$ and $a_2$ such that
$q_1 \oplus \bigoplus_{i=1}^{a_1} \mathbb{H} \cong q_2 \oplus \bigoplus_{i=1}^{a_2} \mathbb{H}$.
The discriminant $d(q)$ is well-defined for a quadratic form $q$ and thus on the Grothendieck-Witt ring by multiplicativity: $d([q_1]-[q_2]) = d(q_1)/d(q_2)$. But in a field $K$ in which $-1$ is not a square -- e.g. $\mathbb{Q}_p$ when $p \not \equiv 1 \pmod{4}$ -- Witt equivalent forms need not have the same discriminant, since $d(\mathbb{H}) = -1 \neq 1 = d(\mathbb{H} \oplus \mathbb{H}) \in K^{\times}/K^{\times 2}$. The signed discriminant remedies this.
(None of this has anything to do with me, of course: you would find the same discussion in any introductory treatment of the algebraic theory of quadratic forms. My writeup is heavily indebted to T.Y. Lam's text.)