If we take a knot $K$ in $S^3$, there are several ways to construct the associated Seifert surface. One way, which I am not familiar with, I just came across in a paper I am reading. It goes like this:

Consider the regular neighborhood $n_K$ of $K$ in $S^3$, which is diffeomorphic to $K\times \mathbb{R}^2$. $S^3-n_K$ is the knot complement. By Alexander duality, $H^1(S^3-n_K)\cong H_1(K)\cong\mathbb{Z}$; but we also have $H^1(S^3-K)\cong \[S^3-n_K,S^1\]$, so we can take a (smooth) map $f:\ S^3-n_K\rightarrow S^1$ representing a generator of $H^1$. The preimage of a regular value of $f$ gives a codimension-1 submanifold of $S^3-n_K$; namely, a surface-with-boundary $S$, such that $\partial S\subset \partial (S^3-n_K)$. So far, so good. It is the next step that throws me:

$\partial S$ is cobordant to $K$.

I don't understand how this works, and am asking how one shows this. For example, it's not even clear that $S$ is connected, and each component could have boundary.

I should mention that this is of course trivial in this specific case, since all 1-manifolds are cobordant. But this should work for higher-dimensional knots. It is even used in this answer and the comments below, where $K$ is replaced by an orientable 3-manifold and $S^3$ is replaced by $S^5$ (or $S^6$).

Can someone enlighten me in how one shows this (in the general case)? Namely, without relying on the dimensions of the spaces involved, but rather their codimensions?

[It might be relevant to note that the "regular neighborhood" (normal bundle) of the two examples above - the knot and the 3-manifold - are trivial. I don't know if this matters or not.]