Let $k$ be a nonnegative integer. Let $K$ be a commutative ring, and let $V$ and $W$ be two $K$-modules. Let $\alpha : V \times W \to K$ be a $K$-bilinear form. Then, there is a $K$-bilinear form
\begin{align}
\alpha_k : \wedge^k V \times \wedge^k W &\to K; \\
\left(v_1 \wedge v_2 \wedge \cdots \wedge v_k , w_1 \wedge w_2 \wedge \cdots \wedge w_k\right) &\mapsto \det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right)
\end{align}
(where $\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}$ is the $k\times k$-matrix with entries $\alpha\left(v_i, w_j\right)$).
Note that the image of $\left(v_1 \wedge v_2 \wedge \cdots \wedge v_k , w_1 \wedge w_2 \wedge \cdots \wedge w_k\right)$ under this form is often called a Gram determinant (though the name traditionally stands for some particular cases), and can be rewritten as $\sum_{\sigma \in S_k} \left(-1\right)^{\sigma} \prod_{i=1}^k \alpha\left(v_i, w_{\sigma\left(i\right)}\right)$ (where $S_k$ is the symmetric group on the set $\left\{1,2,\ldots,k\right\}$, and where $\left(-1\right)^{\sigma}$ denotes the sign of a permutation $\sigma$).

Why is this form $\alpha_k$ well-defined? Here is the straightforward way to see this (there might be slicker arguments): For each $v_1, v_2, \ldots, v_k \in V$, we define a $K$-linear map
\begin{align}
A_{v_1, v_2, \ldots, v_k} : \wedge^k W &\to K; \\
w_1 \wedge w_2 \wedge \cdots \wedge w_k &\mapsto \det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right) .
\end{align}
This is well-defined (by the universal property of $\wedge^k W$), since the determinant $\det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right)$ depends $K$-multilinearly on the $w_1, w_2, \ldots, w_k$ and vanishes when two of $w_1, w_2, \ldots, w_k$ are equal (indeed, this follows from standard properties of determinants). Now, consider the map
\begin{align}
A : V^{\times k} &\to \operatorname{Hom}_K\left(\wedge^k W, K\right); \\
\left(v_1, v_2, \ldots, v_k\right) &\mapsto A_{v_1, v_2, \ldots, v_k} .
\end{align}
This map $A$ is $K$-multilinear and kills all $k$-tuples $\left(v_1, v_2, \ldots, v_k\right) \in V^{\times k}$ that have two equal entries (indeed, this all boils down to checking identities between $K$-linear maps, such as
\begin{align}
A_{v_1, v_2, \ldots, v_{i-1}, v_i + v'_i, v_{i+1}, \ldots, v_k}
&= A_{v_1, v_2, \ldots, v_{i-1}, v_i, v_{i+1}, \ldots, v_k}
+ A_{v_1, v_2, \ldots, v_{i-1}, v'_i, v_{i+1}, \ldots, v_k} ; \\
A_{v_1, v_2, \ldots, v_{i-1}, \lambda v_i, v_{i+1}, \ldots, v_k}
&= \lambda A_{v_1, v_2, \ldots, v_{i-1}, v_i, v_{i+1}, \ldots, v_k} ; \\
A_{v_1, v_2, \ldots, v_{i-1}, v, v, v_{i+2}, \ldots, v_k}
&= 0 ;
\end{align}
but any identity between $K$-linear maps can be proven by evaluating both sides at the generators $w_1 \wedge w_2 \wedge \cdots \wedge w_k$ of $\wedge^k W$; but when evaluated this way, all these identities again follow from basic properties of the determinant). Thus, this map $A$ gives rise to a $K$-linear map
\begin{align}
A' : \wedge^k V &\to \operatorname{Hom}_K\left(\wedge^k W, K\right); \\
v_1 \wedge v_2 \wedge \cdots \wedge v_k &\mapsto A_{v_1, v_2, \ldots, v_k}
\end{align}
(by the universal property of $\wedge^k V$).
This map $A'$, in turn, induces a $K$-bilinear form
\begin{align}
\alpha_k : \wedge^k V \times \wedge^k W &\to K; \\
\left(p, q\right) &\mapsto \left(A'\left(p\right)\right)\left(q\right)
\end{align}
(by uncurrying). An immediate verification reveals that this $K$-bilinear form $\alpha_k$ is precisely the form $\alpha_k$ presented above.

Of course, by considering $k$ as variable, we can glue these $K$-bilinear forms $\alpha_k : \wedge^k V \times \wedge^k W \to K$ together into a $K$-bilinear form $\alpha_\wedge : \wedge V \times \wedge W \to K$. This latter form, I believe, is the form Serre wants.

Why is this latter form $\alpha_\wedge$ non-degenerate when $\alpha$ is non-degenerate? Here, we say that a $K$-bilinear form $\beta : P \times Q \to K$ is non-degenerate if there exist bases $\left(p_i\right)_{i \in I}$ and $\left(q_i\right)_{i \in I}$ of $P$ and $Q$, respectively, such that $\beta\left(p_i, q_j\right) = \delta_{i, j}$ for all $i \in I$ and $j \in I$. Two such bases are called *dual bases* for the form $\beta$.

Now, assume that $\alpha$ is non-degenerate. Thus, there exist dual bases $\left(v_i\right)_{i \in I}$ and $\left(w_i\right)_{i \in I}$ for the form $\alpha$. Consider such bases. WLOG assume that the set $I$ is totally ordered (since we can always equip $I$ with a total order). The basis $\left(v_i\right)_{i \in I}$ of $V$ induces a basis $\left(v_{i_1} \wedge v_{i_2} \wedge \cdots \wedge v_{i_k}\right)_{\left(i_1, i_2, \ldots, i_k\right) \in I_k}$ of $\wedge^k V$, where $I_k$ is the set of all strictly increasing $k$-tuples $\left(i_1 < i_2 < \cdots < i_k\right) \in I^k$. Similarly, the basis $\left(w_i\right)_{i \in I}$ of $W$ induces a basis $\left(w_{i_1} \wedge w_{i_2} \wedge \cdots \wedge w_{i_k}\right)_{\left(i_1, i_2, \ldots, i_k\right) \in I_k}$ of $\wedge^k W$. It is now easy to see that these two bases of $\wedge^k V$ and $\wedge^k W$ are dual bases for the $K$-bilinear form $\alpha_k$. Thus, the $K$-bilinear form $\alpha_k$ has dual bases, i.e., is non-degenerate. Qed.