Quite a nice one is the two-set partition $A,A^c$, where $A$ is the set of points with an even number of terminal 0's:
$$
A =  \bigcup_{k \geq 0} A_k
$$
where $A_k = \bigl\{(x_1, x_2, \ldots) \mid x_1=\ldots=x_{2k}=0 \text{ and } x_{2k+1}=1 \bigr\}$. 
Or, if you work with the odometer acting on the space $[0,1[$:
$$
A = \bigcup_{k \geq 0} \left[\frac{1}{2^{k+1}}, \frac{1}{2^k} \right[ 
$$

This partition is generating because of the following reason. Code a trajectory $x, Tx, T^2x, \ldots$ by a sequence $(v_0, v_1, \ldots)$ of $a$'s and $b$'s according to whether it's in $A$ or not. 

Then you can get the first digit $x_1$ of $x$ by looking at the blocks of two consecutive terms:

* if the first digit of $x$ is $x_1=0$, then the blocks $(v_0,v_1)$, $(v_2, v_3)$, $\ldots$ of two consecutive terms are $aa$ or $ba$ only;
* if it is $x_1=1$, then these blocks are $aa$ or $ab$ only.

To get the second digit of $x$, look at the blocks of four consecutive terms:

* if the first digit of $x$ is $x_1=0$ and:
    - its second digit is $x_2=0$, then the blocks $(v_0,v_1, v_2, v_3)$, $\ldots$ of four consecutive terms are $aaba$ or $baba$ only;
    - its second digit is $x_2=1$, then these blocks are $baaa$ or $baba$ only.

* if the first digit of $x$ is $x_1=1$ and:
    - its second digit is $x_2=0$, then the blocks $(v_0,v_1, v_2, v_3)$, $\ldots$ of four consecutive terms are $abaa$ or $abab$ only;
    - its second digit is $x_2=1$, then these blocks are $aaab$ or $abab$ only.

And so on, looking at the block of $2^n$ consecutive terms provides the $n$-th digit of $x$.