We usually define $V \wedge V := \textrm{span}\{a \otimes b - b \otimes a; \ a, b \in V\}$. If $V$ is a vector space over $\mathbb{K}$, it's obvious that $k(V \wedge V) \cong V \wedge V$ for any nonzero $k \in \mathbb{K}$, so an alternative definition could be $V \wedge V := \textrm{span}\{k(a \otimes b - b \otimes a); \ a, b \in V\}$.
However, it's unclear to me why this definition should be unique. Why wouldn't there be some other "polynomial" in $a \otimes b$ which satisfies the required properties of $\wedge$? For example, a possible definition is $a \wedge b := a \otimes b \otimes a \otimes b - b \otimes a \otimes b \otimes a$. This also satisfies $a \wedge b = -b \wedge a$, although it seems to be nonassociative. Is it possible to find some other similar definition of $\wedge$ that ends up producing an isomorphic space $V \wedge V$?
I'm asking because I haven't been able to find anywhere a derivation of the fact that $a \wedge b = a \otimes b - b \otimes a$ from the more nonexact constructions of $V \wedge V$, such as the standard construction as the quotient of $V \otimes V$ and the space spanned by all tensors of the form $v \otimes v$. Is this because it is impossible to prove that $a \wedge b = a \otimes b - b \otimes a$ from this definition? In that case, there should be other isomorphic spaces which define $\wedge$ differently. If that isn't the case, I'd like to know how to show that this is the unique such space.
This is (up to a multiplicative constant) the only way to do it. The exterior product $a \wedge b$ is also supposed to be bilinear in $a$ and $b$, and this forces us to use tensors of order exactly $2$. So our only option is some linear combination of $a \otimes b$ and $b \otimes a$ and only (scalar multiples of) $a \otimes b - b \otimes a$ are antisymmetric.
More formally, the exterior square $\Lambda^2(V)$ (I think the notation $V \wedge V$ is very bad and should be avoided, because it incorrectly suggests that $V \wedge W$ is meaningful when it's not) embeds in exactly one of the tensor powers $V^{\otimes n}$ in a $GL(V)$-equivariant way, namely the tensor square $V^{\otimes 2}$ (and we can check this by scaling), and this embedding is unique up to scale. In fact we have a canonical decomposition $V^{\otimes 2} \cong S^2(V) \oplus \Lambda^2(V)$ (in characteristic $\neq 2$).