I solved Hatcher's exercise $2B.5$ but I wonder if there is a more elementary approach. Paraphrasing, and removing trivial or vacuous cases, the exercise is this:
Suppose $n\ge1$ and $0\le k\le n-1$ are integers, $h:S^k\hookrightarrow S^n$ and $h':D^n\hookrightarrow S^n$ are embeddings. Write $D^k\subset D^n$ for the subdisk included in the first $k$ coordinates and $D^{n-k}\subseteq D^n$ for the subdisk included in the last $(n-k)$ coordinates.
If $h'(D^n)\cap h(S^k)=h'(D^k)$ then show that the inclusion $h'(\partial D^{n-k})\hookrightarrow S^n\setminus h(S^k)$ induces an isomorphism on homology.
My solution required the general Schoenflies theorem, see here for proofs (I have yet to study this myself).
Denote $\ell:=h'(\frac{1}{2}(D^k\setminus\partial D^k))$ and $D:=h'(D^n\setminus(D^k\setminus\ell))$.
$D\setminus\ell=h'(D^n\setminus D^k)$ strongly deformation retracts onto $h'(\partial D^{n-k})$ so it suffices to show the inclusion $D\setminus\ell\hookrightarrow S^n\setminus h'(S^k)=(S^n\setminus D')\setminus\ell$ induces an isomorphism on homology. By the long exact sequence of a pair and by excision it suffices to show: $$H_n(S^n\setminus D';D)\cong H_n((S^n\setminus D')\setminus\ell;D\setminus\ell)\cong0$$
This follows from the reduced long exact sequence of a pair and the fact that both $S^n\setminus D'$ and $D$ have trivial reduced homology because, by the general Schoenflies theorem, $D'$ is an embedded $k$-disk.
Note that the shortening of $\ell$ to a proper subdisk of $D^k$ so that the intersection $\overline{\ell}\cap h(S^k)$ - the boundary of $D'$ - lies within the 'good space' of $h'(D^n)$ and we can check the manifold/collar neighbourhood condition needed for the general Schoenflies theorem. The shortening of $\ell$ so it lies within $\mathrm{int}\,h'(D^n)$ is also needed for excision to apply.
This is quick and clean and I'm happy with it (well, if you see any errors please let me know!). However I suspect Hatcher had at least one more elementary approach in mind - use of the general Schoenflies theorem should be avoidable. How can we do this?