How does the Cantor's diagonal argument that $(0,1)$ is uncountable deals with the fact some real numbers have two different decimal expansions?

To be a useful bijection, $f$ needs to be $f: \mathbb{N} \rightarrow (0,1)$. Note that, since $f$ is assumed to be bijective, each value of $f(n)$ is unique, regardless of how it is represented.

Assuming we are using base $10$, the $b_{nn}$ can be chosen so they are not equal to $0$ or $9$. Then, no matter how the list is represented, it will not include the new number.

Suppose you consider just the subset of $(0, 1)$ that consists of numbers whose decimal representation does not include an infinite suffix of identical repeating digits. Even this subset cannot be placed into a bijection with the natural numbers, by the diagonal argument, so $(0, 1)$ itself, whose cardinality is at least as large as this subset, must also be uncountable.

You can choose every digit $b_j$ to lie in $\{1,2\}$ say, and so avoid a decimal ending in a string of zeroes or nines.

Alternatively there are countably many decimals ending in all zeroes or all nines, so you can insert them in your original list of decimals. Then your $s$ is also guaranteed no to end in all zeroes or all nines.