I am having trouble understanding TAS. Let $(X,+)$ be a finite abelian group. A translation assiociation scheme is an association scheme $(X, \mathbf{R})$so that for all $(x,y) \in R_i \implies (x+z,y+z) \in R_i \quad \forall i \quad \forall z\in X$.
It is mentioned that if $(X, \mathbf{R})$ is a TAS and $X_i = \{x\in X: (x,0) \in R_i\}$ then $\{X_0,...,X_n\}$ forms a partition of $X$ and $R_i = \{ (x,y) \in X \times X: x-y \in X_i\}$.
I dont really see why or what the point behind this statement is. Is it enough to have a Partition of $X$ (there are many ways to partition $X = \mathbb{Z_6}$ for example) and then the $R_i$ are given then? (because i know one symmetric association scheme with 6 points and 3 classes).
It is said that the Hamming scheme $H(n,q)$ is also a TAS with $X = \mathbb{Z}_q^n$ and partition $X_i = \{ x \in X: wt_H(x) = i\}$ where $wt_H(x)$ counts the number of entries which are not $0$. I have seen already that it is an association scheme but why is it true if only the partition is given (or is there more to it) ?
An association scheme is a partition $\{R_0,\ldots,R_d\}$ of $X \times X$ with certain properties. For a TAS, the partition can alternatively be given as $\{X_0,\ldots,X_d\}$ (with $X_i$ defined as in your question), which is a simpler description as it is a partition of $X$ (and not $X\times X$). The $R_i$ can be recovered as stated in your question.
For a TAS, the derived $R_i$'s must still satisfy the axioms of a translation scheme, of course. So for your first point, not all partitions of $\mathbb{Z}_6$ will give a TAS, but only very specific ones.
For your second point, in the moment that it is understood that the Hamming scheme is a TAS, it is natural to ask what the alternative partition $\{X_0,\ldots,X_d\}$ looks like. This is answered by the statement in your source.