Equivalent condition for a linear map to coincide with another restricted to a line modulo certain hyperplane. Let $p$ be a line of the vector space $K^{n+1}$, and let $H$ be a hyperplane of $K^{n+1}$ such that $p\subseteq H$. We may also interpretate $H$ as a line $q\subseteq(K^{n+1})^{*}$.
Let $\alpha\in\mathrm{Hom}(p,K^{n+1}/p), \beta\in\mathrm{Hom}(H,K^{n+1}/H)$. We can understand $\beta\in\mathrm{Hom}(q,(K^{n+1})^{*}/q)$. Let $v\in p,w\in q$. In p.208 of Harris's 'Algebraic Geometry: A first course' it is said that the condition
$$
\langle\alpha(v),w  \rangle+\langle v,\beta(w)\rangle=0\quad\text{ for } v\in p,w\in q
$$
where $\langle, \rangle$ is the map of the dual pairing, is equivalent to
$$
\beta|_{p}\equiv\alpha \text{ }(\mathrm{mod}\text{ } H).
$$
I don't really understand why this is true. Any hint would be appreciated.
 A: This is not a complete answer to the question, but only my try so far, as requested by the question author. In particular I do not arrive at the given answer, but at one having different signs. As I am not sure about the used identifications I will try to make all identifications explicit, albeit at the costs of elegance.


We may also interpret $H$ as a line $q \subseteq (K^{n+1})^*$.

I assume that we are doing so by taking the orthogonal complement of $H$, i.e.
$$
 q = \{w \in (K^{n+1})^* \mid \text{$w(h) = 0$ for every $h \in H$}\}.
$$

Let $\alpha \in \mathrm{Hom}(p,K^{n+1}/p)$ […]

Because $p \subseteq H$ we can identify $q$ with a subspace of $(K^{n+1}/p)^*$ via
$$
 q \to (K^{n+1}/p)^*,
 \quad w \mapsto w'
 \quad\text{where}\quad
 w'(x+p) := w(x)
 \quad\text{for all $x \in K^{n+1}$}
$$
Similary we can identify $q$ with a subspace of $(K^{n+1}/H)^*$ via
$$
 q \to (K^{n+1}/H)^*,
 \quad w \mapsto w''
 \quad\text{where}\quad
 w''(y+H) = w(y)
 \quad\text{for all $y \in K^{n+1}$}.
$$
(This is actually already an isomorphism.)

[…] $\beta \in \mathrm{Hom}(H, K^{n+1}/H)$. We can understand $\beta \in \mathrm{Hom}(q, (K^{n+1})^*/q)$.

I assume we do so in the following way:
We first identify $K^{n+1}/H$ with $q^*$ by using that
$$
 K^{n+1} \to q^*, \quad x \mapsto (w \mapsto w(x))
$$
is surjective with kernel $H$; the resulting isomorphism is give by
$$
 K^{n+1}/H \to q^*, \quad x \mapsto (w \mapsto w''(x)).
$$
This results in an isomorphism $\mathrm{Hom}(H, K^{n+1}/H) \to \mathrm{Hom}(H, q^*)$, under which $\beta_1 := \beta \in \mathrm{Hom}(H, K^{n+1}/H)$ corresponds to $\beta_2 \in \mathrm{Hom}(H, q^*)$ given by
$$
 \beta_2(h)(w) = w''(\beta(h))
$$
Now we use the natural ismorphism
$$
 \Phi \colon \mathrm{Hom}(H, q^*) \to \mathrm{Hom}(q, H^*), \quad \Phi(f)(w)(h) = f(h)(w).
$$
Under $\Phi$ the element $\beta_2 \in \mathrm{Hom}(H, q^*)$ corresponds to $\beta_3 \in \mathrm{Hom}(q, H^*)$ given by
$$
 \beta_3(w)(h)
 = \Phi(\beta_2)(w)(h)
 = \beta_2(h)(w)
 = w''(\beta(h)).
$$
Lastly we identify $(K^{n+1})^*/q$ with $H^*$ via the restriction
$$
 (K^{n+1})^* \to H^*, \quad \phi \mapsto \phi|_H,
$$
which is surjective with kernel $q$. Under the induced isomorphism $\mathrm{Hom}(q, (K^{n+1})^*/q) \to \mathrm{Hom}(q, H^*)$ the preimage of $\beta_3 \in \mathrm{Hom}(q, H^*)$ is given by $\beta_4 \in \mathrm{Hom}(q, (K^{n+1})^*/q)$.


[…] the condition
  $$
 \langle \alpha(v), w \rangle + \langle v, \beta(w) \rangle = 0
 \quad\text{for every $v \in p$, $w \in q$}
$$
  where $\langle \cdot, \cdot \rangle$ is the map of the dual pairing […]

Using the above identifications this condition is expressed by
$$
 \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0
 \quad\text{for every $v \in p$, $w \in q$}
$$
Now let $v \in p$, and let $\alpha(v) = x+p$ and $\beta(v) = y+H$ with $x,y \in K^{n+1}$. Then
$$
 \langle \alpha(v), w' \rangle
 = w'(\alpha(v))
 = w'(x+p)
 = w(x)
$$
and
$$
 \langle v, \beta(w) \rangle
 = \beta_3(w)(v)
 = w''(\beta(v))
 = w''(y+H) = w(y).
$$
Therefore
\begin{align*}
     &\, \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0 \quad\text{for every $w \in q$} \\
 \iff&\, w(x) + w(y) = 0 \quad\text{for every $w \in q$} \\
 \iff&\, \langle x+y, w \rangle = 0 \quad\text{for every $w \in q$} \\
 \iff&\, x + y \in H \\
 \iff&\, x \equiv -y \pmod{H} \\
 \iff&\, \alpha(v) \equiv -\beta(v) \pmod{H},
\end{align*}
So we have
\begin{align*}
     &\, \langle \alpha(v), w' \rangle + \langle v, \beta_3(w) \rangle = 0 \quad\text{for every $v \in p$, $w \in q$} \\
 \iff&\, \alpha(v) \equiv -\beta(v) \pmod{H} \quad\text{for every $v \in p$} \\
 \iff&\, \alpha \equiv -\beta|_p \pmod{H}.
\end{align*}

While this is not the desired result I hope that I helps and that I just overlooked a sign somewhere. 
