Relation between quadratic refinement and quadratic form The question in the title has now been bothering me for days. I first came across the term quadratic refinement when I read about the Kervaire invariant when reading Kervaire's 1960 paper. The Wikipedia article about quadratic refinement redirects to an article about $\varepsilon$-quadratic form. This seems to suggest that a quadratic refinement and an $\varepsilon$-quadratic form are one and the same thing. Reading the article however seems to suggest another possibility:  that an $\varepsilon$-quadratic form is a generalization of quadratic forms. (Recall: a quadratic form is a particular sort of polynomial defined on a field so that a generalization would be something like a polynomial on a ring or similar.) Finally looking it up in an alternative source I ended up with a definition without $\varepsilon$-quadratic forms. 
I hope to have given an at least somewhat intelligible summary of my work done so far. Now I would need your help with the following question: what is the rôle of quadratic refinements in Kervaire's paper (link above) and ensuing papers and what is the relation between quadratic refinements and quadratic forms? 
 A: I can give you some intuition in the case of surfaces (i.e. : a $2$-dimensional compact manifold without boundary).
Let $V$ be a finite dimensional $k$-vector space, where $k$ is any old field, for now.  When char$(k) \ne 2$, it is not hard to convince oneself that quadratic forms on $V$ and bilinear forms over $V$ are really the same thing; if $B$ is a bilinear form on $V \times V$, $B(x,x)$ is a quadratic form on $V$, and if $q$ is a quadratic form on $V$, $(q(x+y)-q(x)-q(y))/2$ is a bilinear form on $V \times V$.
When char$(k) = 2$, this is no longer true.  When it does happen that a bilinear form is induced by a quadratic form in the above sense in characteristic 2, we say that the quadratic form is a quadratic refinement of the bilinear form.  It appears this is all discussed carefully in the text you referenced.
Now, let's focus on the case where $k=\mathbb{Z}/2\mathbb{Z}$.  Let $V$ be a finite dimensional $\mathbb{Z}/2\mathbb{Z}$ vector space equipped with a non-degenerate, symmetric bilinear form $B:V \times V \to \mathbb{Z}/2\mathbb{Z}$.  Then we can think of $B$ as mapping into $\mathbb{Z}/4\mathbb{Z}$ by composing $B$ with multiplication by $2$; call this new bilinear form $B'$.  It turns out that $B'$ can be "refined" by possibly several quadratic polynomials $q:V \to \mathbb{Z}/4\mathbb{Z}$; that is, there is always at least one quadratic polynomial $q: V \to \mathbb{Z}/4\mathbb{Z}$ such that $B'(x,y)=q(x+y)-q(x)-q(y)$.  When this is the case, we say $q$ is a $\mathbb{Z}/4\mathbb{Z}$ quadratic refinement of $B$.  
Now, what does this have to do with topology?  Let $\Sigma$ be a surface.  It turns out that we can associate to $\Sigma$ a $\mathbb{Z}/2\mathbb{Z}$-vector space $V$ with elements the cobordism classes of immersions $f:Z \looparrowright \Sigma$, where $Z$ is a compact $1$-manifold without boundary (i.e. a disjoint union of circles), and addition given by disjoint union; that is $[Z, f] + [Y, g]=[Z \sqcup Y, f \sqcup g]$ (it turns out that this operation is well-defined, associative, etc.).  The Mod 2 Intersection Form on $\Sigma$ gives a non-degenerate, symmetric bilinear form on $V$ (and, again, it turns out this is well-defined on cobordism classes).
Before the punchline, we need one more definition: say $(V,B)$ is refined by $q_1$ and $q_2$.  We say $q_1$ and $q_2$ are isometric if there exists an automorphism $T$ on $V$ such that $q_1(v)=q_2(T(v))$ for all $v \in V$.
Now, here is the punchline: it turns out that isometry classes of $\mathbb{Z}/4\mathbb{Z}$ quadratic refinements of $V$ correspond bijectively with isotopy classes of immersions of $\Sigma$ into $\mathbb{R}^3$.  Further, one can associate an invariant to a $\mathbb{Z}/4\mathbb{Z}$ quadratic refinement $q$ of $(V,B)$, when $V \ne 0$: $$\kappa:=\frac{1}{\sqrt{V}} \sum _{v \in V} \exp(\frac{\pi}{2} i q(v)).$$  You can guess what $\kappa$ stands for.  One can prove that two quadratic refinements of $V$ are isometric if and only if their $\kappa$ invariants are equal.  
If you prefer, you could also think of $\mathbb{Z}/4\mathbb{Z}$ quadratic refinements of the natural bilinear form $H^1(\Sigma; \mathbb{Z}/2\mathbb{Z}) \times H^1(\Sigma; \mathbb{Z}/2\mathbb{Z}) \to H^2(\Sigma; \mathbb{Z}/2\mathbb{Z})$; this amounts to the same thing, when $\Sigma$ is orientable.  
Here is a paper which references most of what I just said, and discusses generalizations of these ideas to higher dimensional manifolds: http://math.berkeley.edu/~kirby/papers/Kirby%20and%20Taylor%20-%20Pin%20structures%20on%20low-dimensional%20manifolds%20-%20MR1171915.pdf
