Tensor Product is distributive. 
Tensor Product is distributive.

I get stuck at the proof:
If $T$ is a $p$-tensor and $S, U$ a $q$ tensor. Then I need to show that
$$ T \otimes (S \cdot U) = (T \otimes U) \cdot (S \otimes U).$$
Denote $S(u_1, \ldots, u_q) \cdot U(w_1, \ldots, w_q) = S \cdot U (v_1, \ldots, v_q)$. Then
\begin{eqnarray*}
&&T \otimes (S \cdot U)(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q})\\
& =& T (v_1, \ldots, v_p) \cdot (S \cdot U)(v_{p+1}, \ldots, v_{p+q})\\
& =& T (v_1, \ldots, v_p) \cdot (S (u_{p+1}, \ldots, u_{p+q}) \cdot U(w_{p+1}, \ldots, w_{p+q}))\\
&? =& (T \otimes U) \cdot (S \otimes U)
\end{eqnarray*}
 A: Let $V$ be a vector space and $V^*$ the dual space. Suppose $\alpha,\beta \in V^*$. Let $\alpha \otimes \beta :V \times V \rightarrow \mathbb{R}$ be defined by
$$ \alpha \otimes \beta (x,y) = \alpha(x)\beta(y) $$
for all $x,y \in V$. The proof that $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ is as follows: let $x,y \in V$,
$$ \begin{align} \alpha \otimes (\beta + \gamma)(x,y) 
&= \alpha(x)(\beta + \gamma)(y) \\
&= \alpha(x)[\beta(y) + \gamma(y)] \\
&= \alpha(x)\beta(y) + \alpha(x)\gamma(y) \\
&= \alpha \otimes \beta(x,y) + \alpha \otimes\gamma(x,y) \\
&= (\alpha \otimes \beta + \alpha \otimes\gamma)(x,y). \\
\end{align} $$
As $x,y$ were arbitrary the identity  $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ follows. The steps above are justified by the point-wise defined addition of multilinear mappings and the vector space structure of $V$. 
A similar proof can be given for the tensor product of other tensors. I merely give the proof for dual-vectors, or one-forms if you prefer. Hope this helps. I took the liberty of replacing your $\cdot$ with $+$.
