The set of bilinear forms is a right $(R \otimes R)$-module Let $V$ and $A$ be abelian groups. An $A$-valued bilinear form on $V$ is a $\mathbb{Z}$-module homomorphism $$\beta : V \otimes_{\mathbb{Z}} V \rightarrow A$$ Now, let $V$ be a left $R$-module, where $R$ is not necessarily commutative. 
Then, the set of all $A$-valued bilinear forms is a right $(R \otimes R)$-module with the action $$\beta(r \otimes s)(x,y) = \beta(rx,sy)$$ for all $r,s \in R, x,y \in V$.
I'm struggling to see how this is true? How does $\beta(rx,sy)$ even make sense? Isn't $\beta$ just defined on the abelian group $V \otimes_{\mathbb{Z}} V$? How is it then taking values in $V \times V$? Is this $\beta$ supposed to denote the bilinear map $V \times V \rightarrow A$ corresponding to the homomorphism $\beta: V \otimes_{\mathbb{Z}} V \rightarrow A$ by the universal property of the tensor product? If so, how would one prove that the set of these is a right $(R \otimes R)$-module?
 A: In this answer, I explain how $(\beta \cdot (a \otimes b))(x \otimes y) = \beta( (ax) \otimes (by))$ makes sense. Those are equalities (1) and (2) below.

First I want to isolate two claims we will use. These are things we should sort out first if we want to answer the question. For the following two claims, I want to emphasize that I am not using the context of your question (I use the letters differently).

Claim 1: If $M$ is a left $A$-module and $N$ is a left $B$-module, then $M \otimes_{\mathbb{Z}} N$ is a left $A \otimes_{\mathbb{Z}} B$ module. 

We want to set $(a \otimes b) \cdot (m \otimes n) = am \otimes bn$.
$$\text{Hom}_{\mathbb{Z}} ((A \otimes_{\mathbb{Z}} B) \otimes_{\mathbb{Z}} (M \otimes_{\mathbb{Z}} N), M \otimes_{\mathbb{Z}} N) \cong \text{Hom}_{\mathbb{Z}} ((A \otimes_{\mathbb{Z}} M )\otimes_{\mathbb{Z}} (B \otimes_{\mathbb{Z}} N), M \otimes_{\mathbb{Z}} N)$$
This isomorphism comes from the isomorphism $(A \otimes_{\mathbb{Z}} B) \otimes_{\mathbb{Z}} (M \otimes_{\mathbb{Z}} N) \cong (A \otimes_{\mathbb{Z}} M )\otimes_{\mathbb{Z}} (B \otimes_{\mathbb{Z}} N)$, which in turn comes from the commutativity of the tensor product. If $\alpha:  A \otimes_{\mathbb{Z}} M \rightarrow M$ is the structure map for $M$, and $\beta : B \otimes_{\mathbb{Z}} N \rightarrow N$ is the structure map for $N$, then $\alpha \otimes \beta$ is a map  from $(A \otimes_{\mathbb{Z}} M ) \otimes_{\mathbb{Z}} (B \otimes_{\mathbb{Z}} N)$ to $M \otimes_{\mathbb{Z}} N$, and by the isomorphism above, it induces a map from $(A \otimes_{\mathbb{Z}} B) \otimes_{\mathbb{Z}} (M \otimes_{\mathbb{Z}} N)$ to $M \otimes_{\mathbb{Z}} N$. 
Another way to word it is this: Let $\alpha : A \rightarrow \text{End}_{\mathbb{Z}} (M)$ and $\beta : B \rightarrow \text{End}_{\mathbb{Z}}(M)$ be the structure maps. To define a map $\mu: A \otimes_{\mathbb{Z}} B \rightarrow \text{End}_{\mathbb{Z}} (M \otimes_{\mathbb{Z}} N)$, send $a \otimes b$ to the map $\alpha(a) \otimes \alpha(b) : M \otimes_{\mathbb{Z}} N \rightarrow M \otimes_{\mathbb{Z}} N$. For $a, a' \in A$ and $b, b' \in B$, $\mu(a a' \otimes b b') = \mu(a \otimes b) \circ \mu(a' \otimes b')$. 
Both of these constructions result in a multiplication such that $(a \otimes b) \cdot (m \otimes n) = am \otimes bn$. Of course, it's the same multiplication both times. 

Claim 2: If $X$ is a left $A$-module, and a right $B$-module, and $Y$ is a right $B$-module, then $\text{Hom}_{\text{right-} B \text{-mod}}(X,Y)$ is a right $A$-module (note this is reverse from $X$; $X$ is a left $A$-module). 

We put $(\phi \cdot a)(x) = \phi(ax)$. 

Combining these, let's see what we have in your situation. Now I work in the context of your question. $V$ and $A$ are abelian groups. Since $V$ is a left $R$-module, $V \otimes_{\mathbb{Z}} V$ is a left $R \otimes_{\mathbb{Z}} R$-module by (1). So $\text{Hom}_{\mathbb{Z}}(V \otimes_{\mathbb{Z}} V, A)$ is a right $R \otimes_{\mathbb{Z}} R$-module by (2).
Let's calculate what this $R$-module structure on this abelian group. Take $\beta \in \text{Hom}_{\mathbb{Z}}(V \otimes_{\mathbb{Z}} V, A)$, and take $r \otimes s \in R \otimes_{\mathbb{Z}} R$. What is $\beta \cdot (r \otimes s)$?
$$(\beta \cdot (r \otimes s)) (v \otimes w) \stackrel{(1)}{=} \beta (r \otimes s \cdot v \otimes w) \stackrel{(2)}{=} \beta ( (rv) \otimes (sw)) $$
This gives a formula for $\beta \cdot (r \otimes s)$ in terms of $\beta$, $r$ and $s$. The first equality holds by how we defined the $R \otimes_{\mathbb{Z}} R$-module structure on $\text{Hom}_{\mathbb{Z}}(V \otimes_{\mathbb{Z}} V, A)$. The second equality holds because of how we defined the $R \otimes_{\mathbb{Z}} R$-module structure on $V \otimes_{\mathbb{Z}} V$. 
