Given a Banach space $E$, $y\in E$, $\phi\in E^{*}$, I am led to the understanding that $y\odot \phi$ denotes the operator in $B(E)$ defined by $$x\mapsto \phi(x)y,\text{ for all } x\in E$$
Now I have encountered the notation $M\odot N$ for von Neumann algebras $M,N$.
My instinct is to interpret in the following way: I would construct $M\odot N$, the same way I would $M\otimes N$, but with the understanding that $x\otimes y$ is defined to be the operator $x\otimes y\in B(N_{*},M)$ defined by $$\mu\mapsto y(\mu)x,\text{ for all }\mu\in N_{*}$$
Am I close? or way off? Is there a better way to think about this?
Any reference would be appreciated just as much as an answer!
No, that is another pair of shoes. It has nothing to do with the fact that von Neumann algebras are dual spaces.
Short answer: $M\odot N$ denotes the algebraic tensor product of $M$ and $ N$, which has the natural structure of a *-algebra.
You probably know that each von Neumann algebra has a normal, faithful representation on some Hilbert space. Let $\pi\colon M\to B(H), \sigma\colon N\to B(K)$ be such representation. Note that $M\odot N$ has a natural representation $\pi\otimes \sigma$ on $B(H\otimes K)$, where $H\otimes K$ denotes the tensor product of Hilbert spaces, namely $$(\pi\otimes \sigma)(x\otimes y)=\pi(x)\otimes \sigma(y).$$ The tensor product of von Neumann algebras is the second commutant of the range of $\pi\otimes\sigma$.
Note that if $E$ and $F$ are Banach spaces, then you can form their algebraic tensor product $E\odot F$. There is a whole zoo of norms that you can impose on that tensor product.