Let $\{ e_i \}$ be a basis for $V$, then the space of tensors $V \otimes V$ could be identified with the space of all formal sums $\sum_{ij} \alpha_{ij} (e_i, e_j)$ (I know a base independent approach would be preferable, but I want to keep it short).
In the finite-dimensional setting each tensor $T = \sum_{ij} \alpha_{ij} (e_i \times e_j)$ could be identified with a bilinear map $T(u,v) = \sum_{ij} \alpha_{ij} \cdot e_i^*(u) \cdot e_j^*(v)$, where $e_i^*$ denotes the dual basis element defined by $e_i^*(e_j) := \delta_{ij}$, vice versa each bilinear map $T : U \times V \to K$ could be written as a tensor, because the maps $\{ \phi_{ij} \}$ with $\phi_{ij}(u,v) := e_i^*(u)\cdot e_j^*(u,v)$ form a basis of the space of all bilinear maps.
Now for a vector space $V$ set $A(V) := \mbox{span}\{ v \otimes v : v \in V \}$ and define $V \wedge V := V \otimes V / A(V)$. Now I read that $V \wedge V$ could be identified with the space of all alternating forms $V \times V \to K$ (and in many differential geometry books it is defined that way). But how is this identification achieved?
Now I computed an example for $V \wedge V$, let $V = \mathbb R^2$ and let $e_1, e_2$ be the standard bases, then a basis of $V \otimes V$ is $(e_1, e_1), (e_1, e_2), (e_2, e_1), (e_2, e_2)$. And for $v = \alpha e_1 + \beta e_2$ we have $$ v \otimes v = \alpha^2 (e_1, e_1) + \alpha \beta (e_1, e_2) + \beta \alpha (e_2,e_1) + \beta^2 (e_2, e_2) $$ looking at the coordinates, I found that $A(V)$ equals the space where the second and third coordinates equals, i.e. we have $$ A(V) = \mbox{span}\{ (e_1, e_1), (e_2, e_2), (e_1, e_2) + (e_2, e_1) \} $$ and these vector form a basis. With the above interpretation I found that $A(V)$ equals the space of all symmetric bilinear forms.
Now consider $V \wedge V = V \otimes V / A(V)$, then this space is $1$-dimensional, and because $(e_1, e_2) \notin A(V)$ we have $$ V \wedge V = \mbox{span}\{ (e_1, e_2) + A(V) \}. $$
So now in what sense could $V \wedge V$ be identified with the space of all alternating bilinear maps? For example when I try an approach similar as the above, I would identify $\alpha(e_1, e_2) + v$ and $v = \beta(e_1, e_1) + \gamma(e_2, e_2) + \delta( (e_1,e_2) + (e_2, e_1) ) \in A(V)$ with $$ \varphi(u,v) = \alpha e_1^*(u)e_2^*(v) + \beta e_1^*(u)e_1^*(v) + \gamma e_2^*(u)e_2^*(v) + \delta e_1^*(u)e_2^*(v) + \delta e_2^*(u)e_1^*(v). $$ But then not every such element gives an alternating form, for example $(e_1, e_2) + v$ itself, the map $\varphi(u,v) = e_1^*(u) \cdot e_2^*(v)$ is not alternating because $\varphi(e_1, e_2) = 1$, but $\varphi(e_2, e_1) = 0$.
So what went wrong here, and how could this identification achieved?
The main trick here is
$$(x+y) \otimes (x+y) - x \otimes x - y \otimes y = x \otimes y + y \otimes x $$
I bet you're making things much harder by trying to do things in terms of a basis, rather than using tensor algebra.
Also, you're probably contributing to your confusion by identifying elements of $V \otimes V$ with as bilinear forms on $V$: it would be better to identify them with bilinear forms on $V^*$, or consider $V^* \otimes V^*$ instead.