The definition in my book gives it as $\lambda_x$ such that $\lambda_xg=xg$.I understand this much. But my question asks to find the left regular representation of $\mathbb{Z}_3$ in $S_3$.
I don't undertand how to do this at all. If we are working under addition, it doesn't matter whether or not we add from the left or right, correct? I am missing how to connect this to permutations. Any tips/advice would help.
Given any finite group $G$ you can view the left regular representation as a homomorphism $G \rightarrow S_{|G|} \cong Sym(G)$, since left multiplication just permutes group elements.
Therefore, taking $1 \in \mathbb{Z}_3$ ($0$ of course acts as the identity and is rather uninteresting), we have:
\begin{align*}\lambda_1 0 = 1+0=1\\ \lambda_1 1 = 1+1=2\\ \lambda_1 2 = 1+2 =0\end{align*}
So, the action of $1$ on $Sym(G)$ is given by $(0,1,2)$. Can you see how this would generalise for $\lambda_2$/what subgroup of $S_3$ this generates? Hint: it's quite similar to what you began with...
As for the right/left distinction, in this case it doesn't matter since the group is abelian. The left regular representation corresponds to rows (or columns, I can never remember) of the Cayley table, and the right corresponds to columns (or rows). They are of course isomorphic in general, but the distinction is important, since we can get different permutations in $Sym(G)$ by taking the left/right action respectively.