if we have a list of $n$ numbers we need $\log n$ bits
No: if we have a list of numbers between $0$ and $2^k - 1$, we need $k$ bits. There is no relationship between $k$ and $\log n$ in general.
If the numbers are all distinct, then $\log n \ge k$, and radix sort on distinct numbers therefore has a time complexity of $\Omega(n \log n)$. In general, the complexity of radix sort is $\Theta(n \, k)$ where $n$ is the number of elements to sort and $k$ is the number of bits in each element.
To say that the complexity of radix sort is $O(n)$ means taking a fixed bit size for the numbers. This implies that for large enough $n$, there will be many duplicate values.
There is a general theorem that an array or list sorting method that works by comparing two elements at a time cannot run faster than $\Theta(n \log n)$ in the worst case. Radix sort doesn't work by comparing elements, but the same proof method works. Radix sort is a decision process to determine which permutation to apply to the array; there are $n!$ permutations of the array, and radix sort takes binary decisions, i.e. it decides whether to swap two elements or not at each stage. After $m$ binary decisions, radix sort can decide between $2^m$ permutations. To reach the $n!$ possible permutations, it is necessary that $m \ge \log (n!) = \Theta(n \log n)$.
An assumption in the proof that I did not write out above is that the algorithm must work in the case when the elements are distinct. If it is known a priori that the elements are not all distinct, then the number of potential permutations is less than the full $n!$. When sorting $k$-bit numbers, it is only possible to have $n$ distinct elements when $n \le 2^k$; in that case, the complexity of radix sort is indeed $\Omega(n \log n)$. For larger values of $n$, there must be collisions, which explains how radix sort can have a complexity that's less than $\Theta(n \log n)$ when $n \gt 2^k$.