Let $V$ be a an $n$-dimensional vector space over a field $F$. Let $M=F[x]\otimes_F V$. We can consider $M$ as an $F[x]$-module by extending scalars using the inclusion $F\to F[x]$.
Fact 1. There is an $F$-module $\Lambda^k_FM$ and a multi-$F$-linear map $\pi:M^k\to \Lambda^k_F M$ such that all alternating multi-$F$-linear maps defined on $M^k$ factor through $\pi$ to give a unique $F$-linear map from $\Lambda^k_F M$.
Fact 2. There is an $F[x]$-module $\Lambda^k_{F[x]}M$ and a multi-$F[x]$-linear map $\Pi:M^k\to \Lambda^k_{F[x]} M$ such that all alternating multi-$F[x]$-linear maps defined on $M^k$ factor through $\Pi$ to give a unique $F[x]$-linear map from $\Lambda^k_{F[x]}M$.
Note that we are treating $\Lambda^k_F M$ and $\Lambda^k_{F[x]} M$ merely as notations. They may have nothing to do with each other.
Question. Is it true that the gadget $\Pi:M^k\to \Lambda^k_{F[x]}M$ fills the bill for the gadget $\pi:M^k\to \Lambda^k_F M$?
In other words, if suppose we have a multi-$F$-linear map $f:M^k\to N$, then does there exists a unique $F$-linear map $\tilde f:\Lambda^k_{F[x]} M\to N$ whose composition with $\Pi$ is same as $f$?
The motivation for this question is from my discussion with @MooS in this thread.
No, this is not true. For instance, take $n=k=2$, so $M=F[x]^2$, $\Lambda^k_{F[x]}M=F[x]$, and $\Pi:M^2\to F[x]$ is the map that sends $((a,b),(c,d))$ to $ad-bc$ (for $a,b,c,d\in F[x]$). Write $a_n$ for the degree $n$ coefficient of a polynomial $a\in F[x]$. Let $N=F$ and define $f:M^2\to N$ by $f((a,b),(c,d))=a_0d_1-b_1c_0$. Then $f$ is $F$-bilinear and alternating. However, $f$ cannot factor though $\Pi$: for instance, $\Pi((1,1),(x,x))=0$ but $f((1,1),(x,x))=1$.