Consider the map $$ \bigwedge\nolimits^{\!k}(V^*)\to \left( \bigwedge\nolimits^{\!k}V \right)^*\\ \left( \sum_{i=1}^n (f_{1}^i \wedge \dotsb \wedge f_k^i) \right) \mapsto \left( v_1 \wedge \dotsb \wedge v_k\mapsto \sum_{i=1}^n (f_1^i(v_1) \cdot \dotsc \cdot f_k^i(v_k))\right). $$
If $V$ is finite dimensional, this is an isomorphism. Given an element on the right, is there a convenient way to find its inverse image?
I believe you wrote the map backwards and there's a small mistake. Let the isomorphism be $$\varphi : {\bigwedge}^k \bigl( V^* \bigr) \to \left( {\bigwedge}^k V \right)^*.$$
Given $f_1 \wedge \dots \wedge f_k \in \bigwedge^k \bigl( V^* \bigr)$ (IOW each $f_i$ is a functional $V \to \Bbbk$), its image in $\left( \bigwedge^k V \right)^*$ is given by $$\varphi(f_1 \wedge \dots \wedge f_k) : v_1 \wedge \dots \wedge v_k \mapsto f_1(v_1) \cdot \dots \cdot f_k(v_k) \in \Bbbk.$$
(Then you can extend by linearity on sums. Note that what you originally wrote isn't multilinear or symmetric in the $v_i$.)
As usual when a natural map is an isomorphism in finite dimension but not in infinite dimension, you need to pick a basis to explicitly write down the inverse. Let $d = \dim V$ and $(e_1, \dots, e_d)$ be a basis of $V$. Let $(e_1^*, \dots, e_d^*)$ be its dual basis defined by $$e_i^*(e_j) = \delta_{i,j} = \begin{cases} 1 & i = j, \\ 0 & i \neq j. \end{cases}$$
Then if $f \in \left(\bigwedge^k V \right)^*$, its image under $\varphi^{-1}$ is: $$\varphi^{-1}(f) = \sum_{1 \le i_1 < \dots < i_k \le d} \bigl( f(e_{i_1}) e_{i_1}^* \bigr) \wedge \dots \wedge \bigl( f(e_{i_k}) e_{i_k}^* \bigr) \in {\bigwedge}^k V^* $$