You have a couple of misconceptions here.
First, the tensor product is not defined as a quotient of the vector space $V\times W$. Rather, you consider a vector space $Z$ whose basis elements are the elements of $V\times W$. So, for example, if $V=W=\mathbb{R}$, you would have one basis element for $(1,0)$, another basis element for $(2,0)$, another basis element for $(3,0)$, etc. That is why in that post they are written with double brackets, so as not to confuse $Z$ with $V\times W$.
Note that this is way bigger than $V\times W$. The vector space $V\times W$ has dimension $\dim(V)+\dim(W)$. The vector space $Z$ has dimension $|V\times W|$, which is, usually, much larger! For $V=W=\mathbb{R}$, $V\times W$ has dimension $2$, but $Z$ has dimension $\mathfrak{c}=2^{\aleph_0}$.
So you have one basis element for each element of $V\times W$; you should think of $V\times W$ as the index set for the basis. The basis element $[[v,w]]$ is the basis element that corresponds to the element $(v,w)$ of $V\times W$.
Then $E$ is the subspace of $Z$ generated be all the relations you write down; so, in my example above, you would have the vector $2[[1,0]]-[[2,0]]$ in $E$, etc.
Now, the image of the basis vector $[[v,w]]$ in the quotient is denoted by $v\otimes w$. So in general it's not every vector of $Z/E$ that can be written as $v\otimes w$: these are only the images of the basis of $Z$. So you know that these elements generate $Z/E$, but they need not be all of $Z/E$ (in general, they won't be). The elements of $Z/E$ are linear combinations of these "pure tensors" $v\otimes w$.
So, why does it follow from the construction that $(v_1+v_2)\otimes w = (v_1\otimes w) + (v_2\otimes w)$?
This equality is saying that the equivalence class of $[[v_1+v_2,w]]$ is the same as the equivalence class of $[[v_1,w]]+[[v_2,w]]$ in the quotient. By definition of quotient, this is the same as saying that the vector
$$[[v_1+v_2,w]] - [[v_1,w]] - [[v_2,w]]$$
of $Z$ lies in the subspace $E$. But it lies in the subspace $E$ because it is one of the generating elements of $E$. So the equality holds in $Z/E$.
Best Answer
I am guessing that your question is: What is $\bigwedge^k (V \oplus W)$? (This is the wedge product, not the tensor product.)
And also the same question for $S^k = Sym^k$, which is the symmetric product.
Extended hint:
The thing to observe is that there is a natural basis for $V \oplus W$. Namely, you take a basis for $V$ and union it with a basis for $W$.
Let's call our natural basis for $V$ $\{v_1, \ldots, v_n\}$, and our basis for $W$ $\{w_1, \ldots, w_m\}$.
Now, given a basis for a vector space $X$, there is a natural basis for $\bigwedge^k X$ and for $Sym^k X$:
Suppose that $\{x_1, \ldots, x_r\}$ is a basis for $X$.
1) Then a basis for $\bigwedge^k X$ is given by all of the $\{x_{i_1} \wedge \ldots \wedge x_{i_k} \}$ for all $i_1 < \ldots < i_k$, $i_j \in \{1, 2, \ldots r\}$.
2) A natural basis for $Sym$ is given similarly, only now you are allowed to take repeated vectors (so $<$ will be replaced by $\leq$) - another description is the set of monomials of degree k in $\mathbb{Z}[x_1, \ldots, x_r]$.
Finally: Given that $\{x_1, \ldots, x_r\} = \{v_1, \ldots, v_n, w_1, \ldots, w_m\}$ is a basis for $V \oplus W = X$, you can now play a combinatorial game to divide up $\bigwedge^k X$ into direct sums of smaller exterior powers of $V$ and $W$. Similarly for the symmetric product. Do you see how to proceed? Please feel free to ask if you have questions.