Showing bilinearity of the tensor product The tensor product for vector spaces $U,V$ is the initial object in the category of bilinear functionals on $U \times V$.
Spelt out, this means there is a vector space $U \otimes V$, and a bilinear fuctional $i: U \times V \rightarrow U \otimes V$
Such that for any bilinear functional $f: U \times V \rightarrow U \rightarrow W$ to any vector space $W$
There is a unique linear map $f': U \otimes V \rightarrow W$ 
Such that $f=f' \circ g$
Writing $u \otimes v := i(u,v)$, this means $f(u,v)=f'(u\otimes v)$
Given all this, I don't see how we can say that $au\otimes v = u \otimes av = a(u \otimes v)$
edit
The comment below by Yeldarbskich shows I just missed the obvious:
$i: U \times V \rightarrow U \otimes V$  is defined to be bilinear, and renamed to $\otimes$; hence $au \otimes v = i(au,v)=i(u,av)=u \otimes av$
and similar for the other identities.

Take the universal bilinear map $i$ and get your universal $i':U\otimes V\to U\otimes V$. This map is the identity since such a map is unique and the identity satisfies the condition. We have
$$i(au,v)=ai(u,v)=i'(au\otimes v)=ai'(u\otimes v)$$ 
so 
$$au\otimes v=a(u\otimes v)$$ 
Similarly $u\otimes av=a(u\otimes v)$, so we have a three way equality, which is exactly your last set of equations.
