How to construct abstractly and symmetrically the map $𝛬^{n+k}M⊗𝛬^kN→𝛬^nM$ given a pairing $M⊗N→R$?

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.