How to construct abstractly and symmetrically the map $^{n+k}M⊗^kN→^nM$ given a pairing $M⊗N→R$? Let $R$ be a commutative ring and let $M,N$ be $R$-modules.
Let $\langle-,-\rangle$ be a pairing $M⊗N→R$. We can construct a pairing $\Lambda^kM⊗\Lambda^kN→R$ by sending $(a_1∧⋯∧a_k)⊗(b_1∧⋯∧b_k)$ to $\det(\langle a_i, b_j \rangle)_{ij}$.
We can more generally construct a map $\Lambda^{n+k} M ⊗ \Lambda^n N → \Lambda^k M$ that can be defined recursively by
$$ \langle a_1∧⋯∧a_{n+k}, b_1∧⋯∧b_k \rangle = \sum_{i=1}^{n+k} (-1)^{i-1} \langle a_i,b_1\rangle \langle a_1∧⋯∧\widehat{a_i}∧⋯∧a_{n+k}, b_2∧⋯∧b_k \rangle $$
where $\widehat{a_i}$ means that $a_i$ is omitted. The result is a sum over all injections $f : [\![1,k]\!] → [\![1,n+k]\!]$ of $$± \bigg(\prod_{i=1}^k \langle a_{f(i)}, b_i \rangle \bigg) \bigwedge_{j \not\in \operatorname{image}(f)} a_j.$$
Bourbaki gives an abstract construction of these maps as follows, but it is not symmetrical and I don't know how to prove the symmetry of the definition without referring to the formula. Here is the construction of Bourbaki (Algèbre III):
We start by defining a coalgebra structure on $\Lambda M$ by applying $\Lambda$ to the coalgebra structure on $M$ given by the diagonal $M→M⊕M$ and the zero map $M→0$. We get a comultiplication $\Lambda M → \Lambda(M⊕M) = \Lambda M ⊗ \Lambda M$ and a counit $\Lambda M→0=R$. Dualizing degreewise this structure, we get an algebra structure on the graded dual
$$(\Lambda M)' = \bigoplus_{k=0}^∞ (\Lambda^k M)^*.$$ This is an anticommutative alternating graded algebra (which can be deduced from the fact that $\Lambda M → \Lambda (M⊕M)$ is a morphism of algebras, without computation), so by the universal property of $\Lambda (M^*)$, the morphism $M^* → (\Lambda M)'$ gives an algebra morphism $\Lambda (M^*) → (\Lambda M)'$ and thus an action on $\Lambda M$.
So this construction gives the pairing $\Lambda^k M ⊗ \Lambda^k M^* → R$ and if we have a morphism $N→M^*$, we can deduce the pairing $\Lambda^k M ⊗ \Lambda^k N → \Lambda^k M ⊗ \Lambda^k M^* → R$.
Is there an other abstract construction that is more symmetrical than Bourbaki's?
PS: I include the tag "category theory" because perhaps people watching this tag know an answer.
 A: Note that  $\wedge M $ is an $R$ algebra with the following universal property: we have the map $m \mapsto m \in \wedge^1(M) \simeq M$, and $m^2=0$ in $\wedge M$. Moreover, whenever we have an $R$ algebra $A$ and an $R$ linear map from $M$ to $A$ such that any $m$ maps to an element of square $0$, then there exists a map... ( a particular case of Clifford algebra, when the quadratic form is $0$). Now, you need to know how to take the graded tensor products of graded commutative algebras. Once you know that, you know what is $\wedge M \otimes \wedge M$, another graded commutative algebra. Now consider the $R$ linear map from $M$ to $\wedge M \otimes \wedge M$, $m \mapsto m\otimes 1 + 1\otimes m$. Note that we have
$$( m\otimes 1 + 1\otimes m)( m\otimes 1 + 1\otimes m)= m\wedge m \otimes 1 + m \otimes m - m \otimes m + 1 \otimes m \wedge m = 0$$
We conclude there exists a morphism of $R$ algebras
$$\wedge M \to (\wedge M )\otimes (\wedge M)\\
m \mapsto m \otimes 1 + 1 \otimes m$$
Let's calculate for instance the image of $m_1 \wedge m_2 \wedge m_3$. We have
$$m_1 \wedge m_2 \wedge m_3\mapsto ( m_1 \otimes 1 + 1 \otimes m_1)( m_2 \otimes 1 + 1 \otimes m_2)(m_3 \otimes 1 + 1 \otimes m_3) = \\=m_1 \wedge m_2 \wedge m_3\otimes 1  +  (m_1 \wedge m_2 \otimes m_3 + m_2 \wedge m_3 \otimes m_1  +  m_3 \wedge m_1 \otimes m_2) + \\+(\cdots )+ 1 \otimes m_1 \wedge m_2 \wedge m_3$$
We can describe the image of an element $m_1 \wedge \cdots \wedge m_l$. The rule of signs is like the one for the Laplace expansion. The important thing is that globally the permutation across the $\otimes$ sign is even. I hope it is clear enough.
Now, the algebra $\wedge M \otimes \wedge M$ is bi-graded. So we have maps from
$$(\wedge M \otimes \wedge M)^{n+k} \to \wedge^k M \otimes \wedge^n M$$   This gets us an $R$ linear  map
$$\wedge^{n+k} M \to \wedge^k M \otimes \wedge^n M$$
From the above it should be clear what this map is. We used the graded tensor product of $\wedge M$ with itself  to justify that the map is OK, and the correct form. Otherwise, one can just take it as a definition. Anyways, this linear map gets us a linear map
$$\wedge^{n+k} M \otimes \wedge^k N \to \wedge^k M \otimes \wedge^n M\otimes \wedge^k N$$
But we have a linear map
$$\wedge^k M \otimes \wedge^k N\to R\\
(m_1 \wedge\ldots\wedge m_k) \otimes (n_1 \wedge\ldots\wedge n_k) \mapsto \det( \langle m_i, n_j\rangle)$$
Now compose and we get our desired map.
We should also mention the following important fact that shows that one can define the interior product inductively. Say we have $\alpha \in \wedge^{n+k_1+ k_n}$, $\beta_i\in \wedge^{k_i}N$, $i=1,2$. Denote the pairing
$ \wedge^{n+k}M \otimes \wedge^k N \to \wedge^n M$ by
$(\alpha, \beta) \mapsto \alpha\Finv \beta$. Then we have
$$\alpha\Finv(\beta_1\wedge \beta_2) = (\alpha\Finv \beta_1) \Finv\beta_2$$
This follows from the Laplace expansion of determinants.
$\bf{Added:}$ We can see that $\Finv \beta$ is the adjoint of $\beta \wedge \cdot $  on the $N$ side.
