How do we know $S^{k−1} \setminus S^{k−2} = B^{k−1}_1 \sqcup B^{k−1}_{2}$ , two $k − 1$ open balls? How do we know $S^{k−1} \setminus S^{k−2} = B^{k−1}_1  \sqcup B^{k−1}_{2}$ , two $k − 1$ open balls?
 A: For $k = 3$, visualise $S^{k-1} = S^2$ as the surface of the Earth, and visualise $S^{k-2} = S^1$ as the equator. Then $S^{k-1} \backslash S^{k-2}$ is the disjoint union of two $(k-1)$-balls: the northern and southern hemispheres.

This generalises for arbitrary $k$. To write down an explicit homeomorphism, think of $S^{k-1}$ as the unit sphere in $\mathbb R^k$, 
$$ S^{k-1} = \{(x_1, \dots, x_n) \in \mathbb R^{k} : x_1^2 + \dots + x_k^2 = 1 \}$$
and think of $S^{k-2}$ as the subspace $$S^{k-2} = \{(x_1, \dots, x_k) \in S^{k-1} : x_k = 0 \}.$$ Then take $$B_1^{k-1} = \{(x_1, \dots, x_k) \in S^{k-1} : x_k > 0 \}, \ \ \ \ B_2^{k-1} = \{(x_1, \dots, x_n) \in S^{k-1} : x_k < 0 \}.$$
If $B^{k-1}$ is the unit ball in $\mathbb R^{k-1}$,
$$ B^{k-1} = \{(x_1, \dots, x_{k-1}) \in \mathbb R^{k-1} : x_1^2 + \dots + x_{k-1}^2 < 1 \}$$
then we have homeomorphisms $B_1^{k-1} \to B^{k-1}$ and $B_2^{k-1} \to B^{k-1}$ given explicitly by
$$ (x_1, \dots, x_{k-1}, x_k) \mapsto (x_1 , \dots, x_{k-1} ).$$
The inverse map is
$$ (x_1, \dots, x_{k-1}) \mapsto (x_1, \dots, x_{k-1} , \pm \sqrt{x_1^2 + \dots + x_{k-1}^2} ),$$
where the plus sign is taken for $B_1^{k-1}$ and the minus sign is taken for $B_2^{k-1}$.
A: I assume you are asking about $S^{k-1}$ and $S^{k-2}$ in the standard position, i.e. with $S^{k-2}$ placed equatorially on $S^{k-1}$. For arbitrary placement of $S^{k-2}$ in $S^{k-1}$, I don't think it's true.
Consider $S^{k-1}\subseteq {\bf R}^k$ (defined as the solution set of $\lVert \bar x\rVert=1$ for the standard Euclidean norm).
Then the equator of $S^{k-1}$ is just the solution set of $x_k=0$, $\lVert \bar x\rVert=1$. Clearly, the connected components of $S^{k-1}\setminus S^{k-2}$ are the solution sets of $\lVert \bar x\rVert=1, x_k>0$ and $\lVert \bar x\rVert=1, x_k<0$. Now, it is easy to see that the orthogonal projection of each onto the hyperplane $x_k=0$ is the unit ball in the hyperplane, and the projection is a homeomorphism.
