Here is an elementary motivation for interior products .
Suppose $V$ is an $n$-dimensional vector space .
To every non-zero vector $ v\in V$ you can associate the complex $$ 0\to V\to...\to \Lambda ^kV\to \Lambda ^{k+1}V\to...\to \Lambda ^nV\to 0 \quad (\star)$$
where the linear map $ext_k(v):\Lambda ^kV\to \Lambda ^{k+1}V$ is the map $\omega \mapsto v\wedge \omega$.
But how do you prove that it is exact? Answer: with exterior interior products!
Choose a linear form $f\in V^*$ such that $f(v)=1$ and introduce the interior product map $int_k(f):\Lambda ^kV\to \Lambda ^{k-1}V$ which on a decomposable vector reduces to $$int_k(f)(v_1\wedge...\wedge v_k)=\sum_{j=1}^k (-1)^{j+1}f(v_j)v_1\wedge...\wedge \widehat {v_j}\wedge v_k$$ It is then easy to show the relation $$int_{k+1}(f)\circ ext_k(v)+ext_{k-1}(v)\circ int_k(f)=Id_k: \Lambda ^k(V)\xrightarrow {=} \Lambda ^k(V) $$ which immediately implies that the complex $(\star)$ is exact at the $k$-th slot since $$ext_k(v)(\omega)=0\implies \omega = ext_{k-1}(v)[int_{k-1}(f)(\omega)] $$ Differential geometers make essentially the same calculation in the context of the De Rham complex of differential forms: see Exercise 4 of Chapter 7 (Volume I) in Spivak's wonderful A Comprehensive Introduction to Differential Geometry
NB The above is the toy version of the theory of the Koszul complex.
It can always be souped up to any desired degree of unintelligibility (in my case not much souping up is necessary) .
