Let $T^* S^n$ be the disk cotangent bundle on $S^n$ consisting of covectors with norm less than or equal to $1$. I am confused about the long exact sequence of the pair $(T^*S^n, \partial T^*S^n)$. The degree $n$ portion looks like $$ 0 \rightarrow H_{n}(\partial T^*S^n) \rightarrow H_{n}(T^*S^n) \rightarrow H_{n}(T^*S^n, \partial T^*S^n) \rightarrow H_{n-1}(\partial T^*S^n) \rightarrow 0. $$ The leftmost term is $0$ because it equals $H_{n+1}(T^*S^n, \partial T^*S^n) \cong H^{n-1}(T^*S^n) = 0$. I think the middle map $H_n(T^*S^n) \rightarrow H_{n}(T^*S^n, \partial T^*S^n) \cong H^n(T^*S^n)$ is the intersection form map. The middle two term are isomorphic to $\mathbb{Z}$. Under this identification, I think this map can be identified with multiplication by $-\chi(S^n)$. If $n$ is odd, then $\chi(S^n) = 0$ so that $H_n(\partial T^*S^n) \cong H_{n-1}(\partial T^*S^n) \cong \mathbb{Z}$. But if $n$ is even, then $-\chi(S^n) = -2$.
Furthermore, by Poincaré duality, $A: = H_{n}(\partial T^*S^n) \cong H_{n-1}(\partial T^*S^n)$, so the long exact sequence looks like $$ 0 \rightarrow A \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow A \rightarrow 0 $$ with middle map multiplication by $2$. But this is impossible and I am not sure where the contradiction is.
It's not true that $H_n(\partial D^*S^n)$ is isomorphic to $H_{n-1}(\partial D^*S^n)$ when $n$ is even: the former is zero and the latter is $\mathbb{Z}/2$, which is consistent with the rest of your sequence. For example, when $n=2$ the space $\partial D^*S^2$ is $\mathbb{RP}^3$.
You might want to look up the Gysin sequence, which lets you compute the cohomology of sphere bundles like $\partial D^*S^n$.