= Solution
An $R$-orientation of a rank-$r$ <vector bundle> $E\to B$ is a coherent choice of generator in
$$
H^r(E_b,E_b\setminus\{0\};R)
$$
for every fiber. Equivalently, it is a <Thom class> $u_E\in H^r(D(E),S(E);R)$ restricting to those generators. The <Euler class of a vector bundle> is
$$
e(E)=s^*j^*u_E\in H^r(B;R),
$$
where $j:(D(E),\varnothing)\to(D(E),S(E))$ forgets the subspace and $s$ is the zero section.
The <Thom isomorphism theorem> is
$$
H^q(B;R)\xrightarrow{\sim}H^{q+r}(D(E),S(E);R),
\qquad a\longmapsto\pi^*a\smile u_E.
$$
Insert these isomorphisms into the long exact cohomology sequence of $(D(E),S(E))$ and use the deformation retraction $D(E)\simeq B$. This gives the <Gysin sequence of a sphere bundle>
$$
\cdots\to H^{q-r}(B)
\xrightarrow{\smile e(E)}H^q(B)
\xrightarrow{\pi^*}H^q(S(E))
\to H^{q-r+1}(B)
\xrightarrow{\smile e(E)}H^{q+1}(B)\to\cdots.
$$
To verify the labelled maps, represent $a$ under the Thom isomorphism by $\pi^*a\smile u_E$. Its image in $H^*(D(E))$ pulls back along the zero section to
$$
a\smile s^*j^*u_E=a\smile e(E).
$$
Thus multiplication by the Euler class is exactly the map from relative to absolute cohomology in the pair sequence.
Now let $M^m\subset S^n$ and let $V$ be its tubular neighborhood. The normal bundle has rank $r=n-m$. With $\mathbb F_2$ coefficients it is automatically oriented. Since $n>2m+1$, one has $r>m+1$, so its Euler class lies above the dimension of $M$ and vanishes. The Gysin sequence consequently splits into short exact sequences of vector spaces and gives
$$
H^q(\partial V;\mathbb F_2)
\cong H^q(M;\mathbb F_2)\oplus H^{q-r+1}(M;\mathbb F_2).
$$
On the other hand, <Alexander duality> gives
$$
\widetilde H^q(S^n\setminus M;\mathbb F_2)
\cong\widetilde H_{n-q-1}(M;\mathbb F_2).
$$
Over the field $\mathbb F_2$, the latter is naturally dual to $\widetilde H^{n-q-1}(M;\mathbb F_2)$, which expresses the complement cohomology entirely in terms of that of $M$.
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