Suppose $f\colon S^n\to S^n$ is a (surjective, continuous) map such that $f^{-1}(y)=\{x_1, \dots, x_m\}$ for some $y\in S^n$. I would like to calculate $H_n(S^n, S^n\setminus f^{-1}(y))$ (here $R=\mathbb{Z}$ so I won't write it).
Here is my work so far: since $S^n\simeq S^n/(S^n\setminus \{x\})$, we have $H_n(S^n,S^n\setminus \{x\})=\mathbb{Z}$ by functoriality. Then $$ S^n/(S^n\setminus f^{-1}(y))\simeq \lor_m S^n. $$ Furthermore, we know (from class) that $\tilde H(\lor_m S^n)=\oplus_m \mathbb{Z}$. Then once again by functoriality, we have $$ H_n(S^n,S^n\setminus f^{-1}(y)):=\tilde H (S^n/(S^n\setminus f^{-1}(y)))\cong \tilde{H} (\lor_m S^n)\cong \oplus_m\mathbb{Z}. $$ Is my reasoning correct here? This stuff still feels very foreign to me.
I will point out a couple issues with your argument, and give hints on producing a more rigorous solution for this problem.
You seem to have fallen into two classic traps: 1) it's not always true that $X/A \simeq X$ if $A$ is contractible, and 2) it's not always true that $H_*(X,A) \cong \tilde{H}_*(X/A)$. These statements in general require some kind of "good pair" notion, as I'll describe below.
First, your statement $S^n \simeq S^n /(S^n \setminus \{ x\})$ is incorrect, in fact this quotient is homeomorphic to the Sierpinski space, which is contractible. What IS true is that since $S^n \setminus\{x\}$ is contractible there is a homotopy-equivalence of pairs $(S^n, S^n \setminus \{ x\}) \simeq (S^n, \{y\})$ for any point $y$. Then if $m = 1$ we can say $$H_* (S^n, S^n \setminus \{ x\}) \cong H_*(S^n, \{y\}) \stackrel{\text{definition}}{\cong} \tilde{H}_*(S^n). $$
Secondly, you need to be careful about the statement "$H_n(X,A) \cong \tilde{H}_n(X/A)$". In your last equation you seem to imply this is by definition, but the definitions I'm familiar with say that for a pointed space $(X,x)$ the reduced homology is defined as $\tilde{H}_*(X) = H_*(X,\{x\})$, where $H_*(X,A)$ is the homology of the relative chain complex $C_*(X,A)$. The quotient space $X/A$ has a canonical basepoint, namely the equivalence class $A$, and so by definition $\tilde{H}_*(X/A) = H_*(X/A,A/A)$, but it isn't always true that this latter group is isomorphic to $H_*(X,A)$. The appropriate notion is "good pair" (aka "Absolute Neighbourhood Retract"):
See Hatcher, pages 123 and 133. Typical examples of a good pair are: $X$ is a CW complex and $A$ is a subcomplex; $A$ is a smooth submanifold of $X$ of lower dimension (tubular neighbourhood theorem); another buzzword is "cofibration". The pair $(X,A) = (S^n, S^n\setminus \{x \})$ is not good because the subspace is not closed, and we saw above that $H_n(X,A) \cong \mathbb{Z}$ but $\tilde{H}_n(X/A) = 0$ in this case; on the other hand, the pair $(S^n, \{y \})$ is good.
Now for the hints. For notation's sake let's say $X = S^n \setminus \{x_1, \dots , x_m \}$.
With the above remarks about "good pairs" in mind, start by describing a closed subspace $S\subset S^n$ that $X$ deformation retracts to, so that $(S^n, S)$ is therefore a good pair.
Hint: Describe the homotopy type of $D^n \setminus \{x_1,\dots,x_{m-1}\}$, and relate this space to $X$. Try small values of $m$ first.
Now you could proceed with your original idea, by considering the pair $(S^n, S)$ instead of $(S^n, X)$ and computing the quotient space $S^n/S$, and carefully applying the result I stated above.
If algebra is more your thing you can also consider the long exact sequence of the pair, which is a useful tool for computing relative homology groups in general. We already know $H_*(S^n)$ completely, and if you described the homotopy type of $X$ the way I have in mind it's also clear how to completely compute $H_*(X)$, so it's only a matter of filling in the values of $H_*(S^n, X)$. You'll find out that $H_k(S^n, X) = 0 $ if $k\neq n$, and when $k=n$ there is a short exact sequence:
$$ 0 \to H_n(S^n) \to H_n(S^n, X) \to H_{n-1}(X) \to 0 $$
We know $H_n(S^n) \cong \mathbb{Z}$, and $H_{n-1}(X)$ was hypothetically computed above.
Hint: Compute $H_n(S^n,X)$ with the help of the Splitting Lemma.
Both approaches are valid and have their own advantages/disadvantages, it just depends on your comfort levels. I find the geometric argument is conceptually nice but it's harder to make rigorous and you have to be careful with the "good pairs" business, I personally think the long-exact sequence feels cleaner. In both approaches you still have to find a nice description of the homotopy type of $X$.