I can tell you some slightly higher dimensional places where the (Arf-Kervaire)-Brown invariant shows up.

**Links in $S^3$:**

An oriented link $L$ in $S^3$ is called proper if for any component of $L$ the sum of the linking numbers of that component with all of the other components of $L$ is 0 mod 2. Proper links have an Arf invariant defined by taking any properly embedded oriented planar surface in $S^3 \times I$ that has $L$ at one end and some knot $K$ on the other end. Then $\text{Arf}(L)$ is defined to be $\text{Arf}(K)$, and Robertello showed that this definition does not depend on the choice of planar surface.

By changing the orientation of the planar surface, one sees that the Arf invariant of a link does not change if all of the orientations of the components are changed. However, if you change the orientations of just some of the components, then the Arf invariant can change; for example with the (4,2)-torus link. I mention this because when we generalize the Arf invariant to the Brown invariant, we see what is going on here.

Another example to keep in mind is that the Borromean rings have Arf invariant 0 regardless of the orientations of the components. To see this one can find 4 different planar surfaces, corresponding to different orientation possibilities, and compute the corresponding Arf invariants with those, or with the help of the Brown invariant, we can just do this computation once and then since the sum of the pairwise linking numbers is 0, we will soon see that this implies that the orientations do not matter.

In "Local surgery formulas for quantum invariants and the Arf invariant" by Kirby and Melvin, the authors introduce the Brown invariant of a proper link $\beta(L)$. This is a mod 8 invariant of $L$ that is defined by taking an immersed surface $F$ in $S^3$ with boundary $L$, such that the surface induces the 0-framing on $L$, and taking the Brown invariant of a certain quadratic refinement of the intersection from $f_*: H_1(F; \mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/4\mathbb{Z}$. In the case where $f$ is an embedding and $x \in H_1(F; \mathbb{Z}/2\mathbb{Z})$ can be represented by an embedded curve, then $f_*(x)$ is obtained by looking at a neighborhood of $x$ in $F$ which will be a disjoint union of annuli and Möbius bands, and taking the total linking number of the cores of these bands with the boundaries, where componentwise, the boundary and the core are oriented compatibly.

Then for an oriented link $L$ , $4\text{Arf}(L) + lk(L) = \beta(L)$ (mod 8). This formula explains the dependence of the Arf invariant on the orientation of the components.

Also, the Brown invariant shows up when looking at **characteristic surfaces in 4-manifolds:** (in fact this is used by Kirby and Melvin in seeing that $\beta(L)$ does not depend on the choice of surface $F$)

If $F$ is an orientable characteristic surface in a closed smooth 4-manifold $X$ with $H_1(X;\mathbb{Z}) = 0$, then there is a map $q: H_1(F; \mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/2\mathbb{Z}$ which is quadratic with respect to the intersection form on $H_1(F; \mathbb{Z}/2\mathbb{Z})$ whose Arf invariant does not depend on the choice of surface representing the homology class $[F]$. If $x \in H_1(F; \mathbb{Z}/2\mathbb{Z})$ is represented by an embedded curve, then $q(x)$ is the number of times a framed surface bounding $x$ intersects $F$ mod 2. Then it turns out that $\text{Arf}([F]) = (\sigma(X) - [F]\cdot[F])/8$ mod 2, from which Rochlin's theorem follows.

One might wonder what happens if you consider nonorientable characteristic surfaces $F$ in $X$. Then one can similarly define a map $q : H_1(F; \mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/4\mathbb{Z}$ (that agrees with the previous definition if $F$ is orientable), where we identify $\mathbb{Z}/2\mathbb{Z} \subset \mathbb{Z}/4\mathbb{Z}$) and the Brown invariant of this $q$ satisfies $\sigma(X) = [F]\cdot[F] + 2 \beta(F)$ mod 16.

Complete intersections and the Kervaire invariantby W. Browder, in Algebraic topology, Aarhus 1978 — Springer Lecture Notes 763 (1979), 88-108. $\endgroup$ – abx Dec 21 '16 at 6:51