Let $V$ be an $n$-dimensional real vector space and $\omega \in \Lambda^kV^*$ a (decomposable) $k$-covector. Furthermore let $v = v_1\wedge\dots\wedge v_l \in \Lambda^lV$ be any $l$-vector. Define the interior multiplication $i_v: \Lambda^k V^* \to \Lambda^{k-l} V^*$ with the $l$-vector $v$ as the linear map such that $$ i_v \omega = (i_{v_l} \circ \dots \circ i_{v_1}) \omega. $$ where $i_{v_i}$ is the usual interior product. If I take $\omega$ decomposable, i.e. $\omega = f^1\wedge\dots\wedge f^k$, where $f^i \in V^*$ and $k \ge l$, then I have this formula, which I ask about its correctness \begin{align} i_{v_1\wedge\dots\wedge v_l} (f^1\wedge\dots\wedge f^k) &= \sum_{\pi\in S(l,k-l)} \text{sgn}\pi f^{i_1}\wedge\dots\wedge f^{i_l} (v_1,\dots, v_l) f^{j_1}\wedge\dots\wedge f^{j_{k-l}} \\ &= \epsilon_{I,J} f^{I} (v_1,\dots, v_l) f^{J} \end{align} where $S(l,k-l)$ is the set of shuffles, i.e. $\pi: (1,\dots,k) \mapsto (i_1,\dots,i_l,j_1,\dots, j_{k-l}) \equiv (I,J)$ such that $i_1<\dots< i_l$, and $j_1<\dots< j_{k-l}$, and $\text{sgn}\pi$ is the sign of the permutation. The wedge product is defined here as $$ \omega \wedge \eta = \sum_{\pi\in S(k,l)} \text{sgn}\pi(\omega \otimes \eta) \circ \pi= \frac{1}{k! l!} \sum_{\sigma \in S_{k+l}} (-1)^\sigma (\omega \otimes \eta) \circ \sigma. $$
Update:
The way I arrived to this formula is by writing any permutation $\sigma \in S_k$ as product of a shuffle followed by two permutations, each of which permutes one of the sets of indices ($i$s or with $j$s). I denote such factorisation as $$ \sigma = \pi_{\kappa,\rho}:(1,...,k) \mapsto (i_{\kappa 1},\dots,i_{\kappa l},j_{\rho 1},\dots, j_{\rho(k-l)}) $$ where $\pi$ is a shuffle as abobe (see the quoted part in this and the suggested generalisation).
The idea of writing any permutation as a shuffle followed by two permutations, each of which ...(as above), I justify it as follows, its purpose is to group the terms in the long sum in a smart way, namely, in each group we have only terms whose first $l$ factors (in the tensor product) are same up to a permutation (also the last $k-l$ terms are the same up to permutation.) The terms in each group then arise from the same shuffle followed by two permutations. Therefore each group, upon summing over the last two permutations (following the shuffle) gives wedge products. And it remains to sum over the shuffles, and that's the remaining sum in the formula
With this I rewrite the definition \begin{align} i_{v_1\wedge\dots\wedge v_l} (f^1\wedge\dots\wedge f^k) &=i_{v_l}\circ\dots\circ i_{v_1} \left(\sum_{\sigma \in S_k} \text{sgn} \sigma f^{\sigma 1}\otimes\dots\otimes f^{\sigma k}\right)\\ &= \sum_{\sigma \in S_k} \text{sgn} \sigma f^{\sigma 1}(v_1)\dots f^{\sigma l}(v_l)f^{\sigma (l+1)}\otimes\dots\otimes f^{\sigma k} \\ &= \sum_{ \substack{ \pi \in S(l,k-l),\\ \kappa\in S_l,\\ \rho\in S_{k-l} } } \text{sgn} \pi\ \text{sgn} \kappa f^{i_{\kappa1}}(v_1)\dots f^{i_{\kappa l}}(v_l)\ \text{sgn} \rho f^{j_{\rho 1}}\otimes\dots\otimes f^{j_{\rho(k-l)}}\\ &= \sum_{\pi \in S(l,k-l)} \text{sgn}\pi (f^{i_1}\wedge\dots\wedge f^{i_l})(v_1,\dots, v_l) f^{j_1}\wedge\dots\wedge f^{j_{k-l}} \end{align}
I'm not familiar with some of your notation, but your formula looks right. Another way to write it is $$i_v(\omega)=\sum_{i_1<\cdots<i_l}(-1)^{\sum_{j=1}^l(i_j-j)}\langle v,f^{i_1}\wedge\cdots\wedge f^{i_l}\rangle f^{i_{l+1}}\wedge\cdots\wedge f^{i_k}$$ where $i_{l+1}<\cdots<i_k$ is the complementary ordered $(k-l)$-tuple and the angle brackets denote the scalar product between $\bigwedge^l V$ and $\bigwedge^l V^*$.
For a reference see Greub Multilinear Algebra, 2nd. ed. Problem 5.14.6.