Does multilinearity hold in multiple component in tensor product and wedge product? Let $V$ be a vector space over the field $K.$ Now consider $V \otimes V \otimes V \otimes V$ and for $x_i \in V$ let $x_1 \otimes x_2\otimes (x_3\otimes x_4 + x_5 \otimes x_6)$ be an element in the 4-tensor of $V$ . Is it same as $x_1 \otimes x_2\otimes x_3\otimes x_4  + x_1 \otimes x_2\otimes  x_5 \otimes x_6$ ? I also want to know if they are not equal in the tensor product can they (image in the wedge product) be same in the wedge product ?
 A: They will be the "same" element in the sense there is a natural isomorphism between the spaces involved which will map one to the other. This is quite annoying to write down precisely and depends on your definitions of the tensor product and the meaning of the symbols involved but let me try. 
Let's assume we have defined the tensor product of two vector spaces via the universal property (even if you define it via some specific construction, the underlying problem will remain). Hence, given two vectors spaces $V_1, V_2$, we have a vector space $V_1 \otimes V_2$ and a bilinear map $\otimes_{V_1,V_2} \colon V_1 \times V_2 \rightarrow V_1 \otimes V_2$ which is universal in an appropriate sense. So when $v_1 \in V_1$ and $v_2 \in V_2$, the meaning of the symbol $v_1 \otimes v_2$ is the image of the pair $(v_1,v_2)$ under the map $\otimes_{V_1,V_2}$ which I'll write as $\otimes_{V_1, V_2}(v_1,v_2)$.
Now let's say we have three spaces $V_1,V_2,V_3$. What is the meaning of $V_1 \otimes V_2 \otimes V_3$? We have three options:


*

*Interpret $V_1 \otimes V_2 \otimes V_3$ as $V_1 \otimes (V_2 \otimes V_3)$. Then, when $v_i \in V_i$ we can talk about the element $v_1 \otimes (v_2 \otimes v_3)$ which, more precisely, is the element $\otimes_{V_1,V_2 \otimes V_3}(v_1, \otimes_{V_2,V_3}(v_2,v_3))$.

*Interpret $V_1 \otimes V_2 \otimes V_3$ as $(V_1 \otimes V_2) \otimes V_3$. Then, when $v_i \in V_i$ we can talk about the element $(v_1 \otimes v_2) \otimes v_3$ which, more precisely, is the element $\otimes_{V_1 \otimes V_2, V_3}(\otimes_{V_1,V_2}(v_1,v_2),v_3)$.

*Define the tensor product of three vector spaces $V_1,V_2,V_3$ via the universal property so that you have a vector space $V_1 \otimes V_2 \otimes V_3$ and a trilinear map $$\otimes_{V_1,V_2,V_3} \colon V_1 \times V_2 \times V_3 \rightarrow V_1 \otimes V_2 \otimes V_3$$ which is universal in an appropariate sense. Then, given $v_i \in V_i$, we have the element $v_1 \otimes v_2 \otimes v_3$ which is more precisely the image $\otimes_{V_1,V_2,V_3}(v_1,v_2,v_3)$ of the triple $(v_1,v_2,v_3)$ under the map $\otimes_{V_1,V_2,V_3}$.


Now, a priori the elements $v_1 \otimes (v_2 \otimes v_3), (v_1 \otimes v_2) \otimes v_3$ and $v_1 \otimes v_2 \otimes v_3$ live in different vector spaces. Even if you would have defined the tensor product via some specific construction, this would probably still be the case so the statement that $v_1 \otimes (v_2 \otimes v_3) = v_1 \otimes v_2 \otimes v_3$ strictly doesn't make any sense. However, what is true is that there are natural isomorphisms between each of the three constructions which allows us to think of elements living in one space as elements living in any of the other two spaces. This is what is meant by the associativity of the tensor product.
How is this related to your question? Let's remove one copy of $V$. Given $v_1,\dots,v_5 \in V$, you are asking whether 
$$v_1 \otimes (v_2 \otimes v_3 + v_4 \otimes v_5) = v_1 \otimes v_2 \otimes v_3  + v_1 \otimes v_4 \otimes v_5. $$
Let's write this more precisely. The first element, as written, is naturally an element of $V \otimes (V \otimes V)$ given by
$$ \otimes_{V,V \otimes V}(v_1, \otimes_{V,V}(v_2,v_3) +_{V \otimes V} \otimes_{V,V}(v_4,v_5)). $$
The second element, as written is naturally an element of $V \otimes V \otimes V$ given by
$$ \otimes_{V,V,V}(v_1,v_2,v_3) +_{V \otimes V \otimes V} \otimes_{V,V,V}(v_1,v_4,v_5). $$
Those elements live in different vector spaces but the map $V \otimes V \otimes V \rightarrow V \otimes (V \otimes V)$ given by $v_1 \otimes v_2 \otimes v_3 \mapsto v_1 \otimes (v_2 \otimes v_3)$ (which exists by the universal property) allows you to identify them and under this identification, your identity will hold.
