Let us denote $N^\vee:=\text{Hom}_A(N,A)$, $S^\perp:=\{f\in N^\vee\mid fS=0\}\subset N^\vee$, and $\hat S:=S^{\perp\perp}\subset N^{\vee\vee}$ for any $A$-module $N$ and its submodule $S\subset N$. If $N$ is finitely generated free, then $N^{\vee\vee}=N$, $S\subset\hat S$, and $S^\perp=\hat S^\perp$.

In these terms, $\alpha\in F^\vee\otimes_AF^\vee$ is antisymmetric (i.e., $\tau\alpha=-\alpha$) with $\alpha(M,M)=0$ and $\beta\in M^\perp\otimes_AF^\vee\subset F^\vee\otimes_AF^\vee$ (i.e., $\beta(M,F)=0$) so that $\beta^*=\tau\beta$, where $\tau$ permutes the $F^\vee$'s in $F^\vee\otimes_AF^\vee$.

There is a tautological criterion: $\alpha=\beta-\beta^*$ iff $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

$\Longrightarrow$ is immediate.

$\Longleftarrow$ if $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$ is antisimmetric, then $\alpha=\beta_1+\tau\beta_2$ for some $\beta_1,\beta_2\in M^\perp\otimes_AF^\vee$ and $\beta_1+\tau\beta_2=-\tau\beta_1-\beta_2$. So, $\alpha=\beta-\beta^*$, where $\beta:=\frac12(\beta_1-\beta_2)$.

In particular, $\alpha(\hat M,\hat M)=0$ is a necessary condition.

Most practical can be the sufficient condition $(M\otimes_AM)^\perp=M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

Let us denote $N^\vee:=\text{Hom}_A(N,A)$, $S^\perp:=\{f\in N^\vee\mid fS=0\}\subset N^\vee$, and $\hat S:=S^{\perp\perp}\subset N^{\vee\vee}$ for any $A$-module $N$ and its submodule $S\subset N$. If $N$ is finitely generated free, then $N^{\vee\vee}=N$, $S\subset\hat S$, and $S^\perp=\hat S^\perp$.

In these terms, $\alpha\in F^\vee\otimes_AF^\vee$ is antisymmetric (i.e., $\tau\alpha=-\alpha$) with $\alpha(M,M)=0$ and $\beta\in M^\perp\otimes_AF^\vee\subset F^\vee\otimes_AF^\vee$ (i.e., $\beta(M,F)=0$) so that $\beta^*=\tau\beta$, where $\tau$ permutes the $F^\vee$'s in $F^\vee\otimes_AF^\vee$.

There is a tautological criterion: $\alpha=\beta-\beta^*$ iff $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

$\Longrightarrow$ is immediate.

$\Longleftarrow$ if $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$ is antisymmetric, then $\alpha=\beta_1+\tau\beta_2$ for some $\beta_1,\beta_2\in M^\perp\otimes_AF^\vee$ and $\beta_1+\tau\beta_2=-\tau\beta_1-\beta_2$. So, $\alpha=\beta-\beta^*$, where $\beta:=\frac12(\beta_1-\beta_2)$.

In particular, $\alpha(\hat M,\hat M)=0$ is a necessary condition.

Most practical can be the sufficient condition $(M\otimes_AM)^\perp=M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

Let us denote $N^\vee:=\text{Hom}_A(N,A)$, $S^\perp:=\{f\in N^\vee\mid fS=0\}\subset N^\vee$, and $\hat S:=S^{\perp\perp}\subset N^{\vee\vee}$ for any $A$-module $N$ and its submodule $S\subset N$. If $N$ is finitely generated free, then $N^{\vee\vee}=N$, $S\subset\hat S$, and $S^\perp=\hat S^\perp$.

In these terms, $\alpha\in F^\vee\otimes_AF^\vee$ is antisymmetric (i.e., $\tau\alpha=-\alpha$) with $\alpha(M,M)=0$ and $\beta\in M^\perp\otimes_AF^\vee\subset F^\vee\otimes_AF^\vee$ with $\beta^*=\tau\beta$, where $\tau$ permutes the $F^\vee$'s in $F^\vee\otimes_AF^\vee$.

There is a tautological criterion: $\alpha=\beta-\beta^*$ iff $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

$\Longrightarrow$ is immediate.

$\Longleftarrow$ if $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$ is antisimmetric, then $\alpha=\beta_1+\tau\beta_2$ for some $\beta_1,\beta_2\in M^\perp\otimes_AF^\vee$ and $\beta_1+\tau\beta_2=-\tau\beta_1-\beta_2$. So, $\alpha=\beta-\beta^*$, where $\beta:=\frac12(\beta_1-\beta_2)$.

In particular, $\alpha(\hat M,\hat M)=0$ is a necessary condition.

Most practical can be the sufficient condition $(M\otimes_AM)^\perp=M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

Let us denote $N^\vee:=\text{Hom}_A(N,A)$, $S^\perp:=\{f\in N^\vee\mid fS=0\}\subset N^\vee$, and $\hat S:=S^{\perp\perp}\subset N^{\vee\vee}$ for any $A$-module $N$ and its submodule $S\subset N$. If $N$ is finitely generated free, then $N^{\vee\vee}=N$, $S\subset\hat S$, and $S^\perp=\hat S^\perp$.

In these terms, $\alpha\in F^\vee\otimes_AF^\vee$ is antisymmetric (i.e., $\tau\alpha=-\alpha$) with $\alpha(M,M)=0$ and $\beta\in M^\perp\otimes_AF^\vee\subset F^\vee\otimes_AF^\vee$ (i.e., $\beta(M,F)=0$) so that $\beta^*=\tau\beta$, where $\tau$ permutes the $F^\vee$'s in $F^\vee\otimes_AF^\vee$.

There is a tautological criterion: $\alpha=\beta-\beta^*$ iff $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.

$\Longrightarrow$ is immediate.

$\Longleftarrow$ if $\alpha\in M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$ is antisimmetric, then $\alpha=\beta_1+\tau\beta_2$ for some $\beta_1,\beta_2\in M^\perp\otimes_AF^\vee$ and $\beta_1+\tau\beta_2=-\tau\beta_1-\beta_2$. So, $\alpha=\beta-\beta^*$, where $\beta:=\frac12(\beta_1-\beta_2)$.

In particular, $\alpha(\hat M,\hat M)=0$ is a necessary condition.

Most practical can be the sufficient condition $(M\otimes_AM)^\perp=M^\perp\otimes_AF^\vee+F^\vee\otimes_AM^\perp$.