Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a $(R,S)$-bimodule. I am trying to show that $M$ is a left $R \otimes_k S^{\text{op}}$-module.
Define a map $\phi \colon R \times S^{\text{op}} \to M$ given by $\phi(r,s) = (rm)s$. I believe that for me to progress with the proof this map has to have the property that $\phi (ar,s) = \phi (r, as)$ for $a \in k$. However $\phi(ar,s) = [(ar)m]s$ and $\phi(r, as) = (rm)(as)$. However I have no idea how to get that $a$ to pass to the other side of the $m$. How is this achieved? Is there implicitly a $k$-module structure on $M$?