Let $R$ be a division ring and let $Z(R)$ be its center. Then clearly $Z(R)$ is a field. It is then claimed that $R$ forms a vector space over its center, but I can't see how that is so.
We know that by Lagrange theorem, $|R/Z(R)|=n$ for some $n\in\mathbb{N}$, and so we can partition $R$ into $n$ cosets, with $Z(R)$ being one of them. So if $R$ a n-dimensional vector space over $Z(R)$, then each member of $R$ is a vector with $n$ entries, and each entry is a member of $Z(R)$.
But I don't quite see how such a vector space has the same set as $R$?
In other words, how can you possibly represent every element of the ring $R$ as a vector with $n$ entries over the field $Z(R)$?
None of what you are considering has anything to do with verifying that $R$ is a vector space over $Z(R)$. To verify that, you just have to use the definition of a vector space. That is, $R$ must have an addition operation and a scalar multiplication operation (with "scalars" being elements of $Z(R)$) which satisfy a certain list of axioms. The addition is just the usual ring addition of $R$, and the scalar multiplication is just the usual ring multiplication of $R$, since elements of $Z(R)$ are also elements of $R$. The vector space axioms then follow immediately from the ring axioms.