You are correct that it is an isomorphism, by the argument you referred to. The Wikipedia page on MV-sequence has a pretty nice description of the boundary map, explaining why it is well-defined etc. I guess the most involved part is seeing why you can sub-divide cycles so that they lie entirely in $A$ or $B$, after this it is simple algebra to define the boundary map.
For the sake of visualising the map you can assume you have a nice triangulation of $S^{3}$ by tetrahedra such that the boundary torus (say $T = \partial N(K)$) is contained in the $2$-skeleton. Then we can partition our set of tetrahedra into two subsets, depending on which connected component of $S^{3} \setminus T$ the interior of each tetrahedron is contained in. This gives us our two cycles $U,V$, with
$U+V = [S^{3}]$.
$\partial U = - \partial V = [T]$.
Then, by definition, the boundary map maps $U+V \in H_{3}(S^{3},\mathbb{Z})$ to $\partial U \in H_{2}(T,\mathbb{Z})$, hence it maps fundamental class to fundamental class and so is an isomorphism.
Is your idea the following? If so, then yes this works.
Take the usual cover of the sphere $S^n$ by open neighborhoods of the upper and lower hemispheres, $U$ and $V$. Let $\sigma$ denote the antipodal map. WLOG we can assume that $\sigma U = V$. We also assume that $U\cap V$ deformation retracts onto the equator, $S^{n-1}$.
Then $\sigma$ induces a map of triples $(S^n,U,V)\to (S^n,V,U)$. Thus since $H^n(U)=H^n(V)=H^{n-1}(U)=H^{n-1}(V)=0$ for $n>1$, we get a map of Mayer-Vietoris sequences:
$$
\require{AMScd}
\begin{CD}
0 @>>>H_n (S^n) @>\partial_{S^n,U,V}>> H_{n-1}(S^{n-1}) @>>> 0 \\
@. @V\sigma_* VV @V\sigma_* VV @. \\
0 @>>>H^n (S^n) @>\partial_{S^n,V,U}>> H_{n-1}(S^{n-1}) @>>> 0. \\
\end{CD}
$$
Now the trick is to recall the definition of the boundary map, $\partial_{S^n,U,V}$.
Given a cycle $z \in C_n(S^n)$, find an equivalent cycle of the form $x-y$, with $x$ a cycle in $C_n(U)$, $y$ a cycle of $C_n(V)$, apply the boundary map to the pair $(x,y)$ to get the pair $(\partial x,\partial y)$, then pullback along the inclusion of $C_{n-1}(S^{n-1})\hookrightarrow C_{n-1}(U)\oplus C_{n-1}(V)$. I.e., take the chain $\partial x = \partial y$, which is a chain in $S^{n-1}$.
The only difference when computing the boundary map $\partial_{S^n,V,U}$ is that now we need to have the cycle from $C_n(U)$ be negative. So we write
$z=(-y)-(-x)$, where $x$ and $y$ are the same cycles as before. Then
we apply the boundary map to get the pair $(-\partial y, -\partial x)$, then the final chain is $-\partial x = -\partial y$ regarded as an element of $C_{n-1}(S^{n-1})$.
Thus $\partial_{S^n,U,V} = -\partial_{S^n,V,U}$.
This says that for $n>1$
$$\deg_{S^n}\sigma = -\deg_{S^{n-1}}\sigma.$$
Moreover, since $\deg_{S^1}\sigma=1$, this yields that $\deg_{S^n}\sigma = (-1)^{n+1}$, as desired.
Not sure if this is what you meant, but do let me know.
Best Answer
You said that you are working with Vick's "Homology Theory". In Chapter 1 he introduces singular homology theory and studies its properties.
In the example on p. 22 / 23 he proves that $H_1(S^1) \approx \mathbb Z$ and shows that the $1$-cycle $c + d$ with singular $1$-simplices $c , d : \sigma_1 \to S^1$ as depicted in Fig. 1.6 represents a generator $\omega$ of this group.
For the sake of convenience let $h : I = [0,1] \to \sigma_1, h(t) = (1-t)(1,0) + t(0,1)$, denote the obvious homeomorphism and let $$\bar c : I \to S^1, \bar c(t) = e^{\pi i t}, \\\bar d : I \to S^1, \bar d(t) = e^{\pi i t + \pi} .$$ Then $c = \bar c \circ h^{-1}, d = \bar d \circ h^{-1}$. Note that we regard $S^1$ as the subset of $\mathbb C$ consisting of all $z$ with $\lvert z \rvert = 1$.
Let us next prove the following
Lemma. Let $\bar u , \bar v : I \to X$ be two paths in a space $X$ such that $u(1) = v(0)$ and $$\bar w = \bar u * \bar v : I \to X, t \mapsto \begin{cases} \bar u(2t) & t \le 1/2 \\ \bar v(2t-1) & t \ge 1/2 \end{cases}$$ be the concatenated path. Let $u = \bar u \circ h^{-1}, v = \bar v \circ h^{-1}, w = \bar w \circ h^{-1}$ be the associated singular $1$-simplices. Then the $1$-chain $u + v - w$ (which is actually a $1$-cycle) is the boundary of a singular $2$-simplex in $X$.
Proof. The idea is this: Consider the standard $2$-simplex $\sigma_2$ with vertices $v_0 = (1,0,0), v_1 = (0,1,0), v_2 = (0,0,1)$. Map the line segment $[v_0,v_1]$ by $u$ to $X$, $[v_1,v_2]$ by $v$ to $X$ and $[v_0,v_2]$ by $w$ to $X$. You get a map from the boundary of $\sigma_2$ to $X$. It is easy to see that this map extends to $\sigma_2$ which proves the lemma.
The lemma shows that the $1$-cycles $c + d$ and $e = \bar e \circ h^{-1}$ with
$$\bar e : I \to S^1, \bar e(t) = e^{2\pi i t}$$ represent the same element in $H_1(S^1)$. Actually both represent the generator $\omega$.
For $k = 1, \ldots, m$ define paths $$\bar p_k : I \to S^1, \bar p_k(t) = e^{\frac{2\pi }{m} i t + \frac{2\pi (k-1)}{m} i} .$$
The induced singular $1$-simplices $p_k = \bar p_k \circ h^{-1}$ give a $1$-cycle
$$\gamma = p_1 + \ldots + p_m .$$
Applying the lemma inductively, we see that $\gamma$ and $e$ represent the same element in $H_1(S^1)$.
Using complex numbers, the map $f$ is given by $$f(z) = z^m .$$ We have $$f_*(\gamma) = f \circ p_1 + \ldots + f \circ p_m.$$ An easy computation shows that $$f \circ p_k = e, $$ thus $$f_*(\gamma) = me .$$ Going to homology we get $$f_*(\omega) = f_*([\gamma]) = [f_*(\gamma)] = [me] = m[e] = m\omega .$$ This shows that $f$ has degree $m$.