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 $m$.
Best Answer
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.