We find all recurrent points in $R$ and $L$ and one other type. These are all recurrent points, since all recurrent points are non-wandering, and any non-wandering point is contained in one of these three types. (See How to determine the non-wandering set Ω(T) (if possible at all)?)
First let us look at some neighborhoods of points:
For $k>n$ let the projection $P_{n,k}:X=\{0,1,2\}^{\Bbb{Z}}\to \{0,1,2\}^{k-n+1}$ be given by
$$
P_{n,k}(Y)=(y_{n},y_{n+1},\dots,y_{k-1},y_k)\quad\text{for}\quad Y=(y_n)_{n\in\Bbb{Z}}\in X.
$$
Then for each $n>0$, the set
$$
V_{n,k}(Y):=\{Z\in X\ :\ P_{n,k}(Z)=P_{n,k}(Y)\}
$$
is an open neighborhood of $Y$. We also set $P_n:=P_{-n,n}$ and define correspondingly $V_n$.
Now we see a consequence for recurrent points in
$$
L=\{ \text{Every 1 has a 2 to its right,
every 2 has a 0 to its right,}
$$
$$
\text{every 0 has a 0 or a 1 to its right.}\}
$$ (or by symmetry in $R$).
Let $Y$ be a recurring point in $L$. Then for all $n>0$ there exists $k>0$ such that
$f^k(Y)\in V_n(Y).$ But $f$ is the shift to the left, so this means
that $y_{n+k}=y_n, y_{n+k-1}=y_{n-1},\dots,y_{-n+k}=y_{-n}$, or in other words, that
$P_{-n+k,n+k}(Y)=P_n(Y)$.
Repeating the procedure with $V_{n+k}$ we obtain a $k_2$ such that $P_{-n-k+k_2,n+k+k_2}(Y)=P_{n+k}(Y)$.
Inductively we obtain a sequence of values $k_1=k,k_2,k_3,\dots >0$ such that
$$
P_{-n-(\sum_{i=1}^{j-1}k_i)+k_j,n+\sum_{i=1}^{j}k_i)}(Y)=P_{n+\sum_{i=1}^{j-1}k_i}(Y).
$$
Setting $n_0:=n$ and $n_{j+1}:=n_j+k_j$ we can rephrase the statement: There exists a strictly increasing sequence
$n_0<n_1<n_2<\dots$ such that
$$
P_{-n_{j-1}+k_j,n_{j-1}+k_j}(Y)=P_{n_{j-1}}(Y),
$$
where $k_j:=n_j-n_{j-1}$. It is also straightforward to check that one can take $n_j>3n_{j-1}+4$ (check for example that
$P_{-n_0+(k_1+k_2),n_0+k_1+k_2}(Y)=P_{n_0}(Y)$, which shows that one can take $k_1$ (and similarly $k_i$) as big as you want).
On the other hand, given a sequence $n_0<n_1<n_2<\dots$ with $n_j>3n_{j-1}+4$, it is possible to describe all $Y\in X$ such that
$$
P_{-n_{j-1}+k_j,n_{j-1}+k_j}(Y)=P_{n_{j-1}}(Y),
$$
where $k_j:=n_j-n_{j-1}$, and these elements are non-wandering. The construction goes as follows. Take $(y_{-n_0},\dots,y_{n_0})$ any sequence in $L$.
Then this determines $(y_{n_1-2n_0-1},\dots,y_{n_1})$. Fill the empty spaces as you want such that $(y_{-n_1},\dots,y_{n_1})$ is a sequence in $L$.
Then this determines $(y_{n_2-2n_1-1},\dots,y_{n_2})$. Fill the empty spaces as you want such that $(y_{-n_2},\dots,y_{n_2})$ is a sequence in $L$.
Clearly this proceeding leads to a recurring point.
So we have determined all recurring points in $L$ (and by symmetry in $R$). Note that the recurring points
associated to an increasing sequence could be associated to another sequence as well (for example periodic points).
Now let us determine recurring points that have a sink (See How to determine the non-wandering set Ω(T) (if possible at all)?), take for example
$$
21021021021021021\ 0\ 12012012012012012.
$$
We shall assume that the sink is situated at
$0$ (by translation we can do that). Then the positive part is $P_+(Y_+)$ of an element $Y_+$ of $L$ and the negative part
is $P_-(Y_-)$ of an element $Y_-$ of $R$, where $P_+=(y_0,y_1,\dots)$ and $P_-(Y)=(\dots,y_{-2},y_{-1})$. Now let $Y$ be a recurring point with a sink at $0$
and let be $n>0$. Then there exists $k>0$ such that $V_n(Y)\cap f^k(V_n)\ne \emptyset.$ By a similar argument as before we can assume $k>n+4$, we set $n_0:=n$,
$n_1:=n+k$ and obtain an increasing sequence $n_0<n_1<n_2<,\dots$ such that
$$
P_{-n_j,-n_j+n_{j-1}-1}(Y)=P_{-n_{j-1},-1}(Y)\quad\text{and}\quad P_{0,n_{j-1}}(Y)=P_{n_j-n_{j-1},n_j}(Y).
$$
As before, we can construct for each increasing sequence $n_0<n_1<n_2<\dots$ with $n_j>3n_{j-1}+4$, all $Y\in X$ such that
$$
P_{-n_j,-n_j+n_{j-1}-1}(Y)=P_{-n_{j-1},-1}(Y)\quad\text{and}\quad P_{0,n_{j-1}}(Y)=P_{n_j-n_{j-1},n_j}(Y),
$$
and these are all recurring points with a sink at the origin.
Note that we don't need that $Y$ be symmetric, we only need that the sink returns after $n_j-n_{j-1}$ steps to the origin.
