Big O: Analyzing the time complexity of an $O(n \log n)$-algorithm
Let's denote by $C(n)$ the number of comparisons performed on an input of length $n$ (later on I will comment more on this definition). The assumption that the number of comparisons roughly doubles when doubling the input size translates to $C(2n) \approx 2 C(n)$. This is not a precise statement, so let us take it to mean that
$$
\lim_{n\to\infty} \frac{C(2n)}{C(n)} = 2.
$$
There are many functions satisfying this statement. Some examples are $n (\log n)^k$, for any value of $k$. This condition cannot distinguish between algorithms with a linear number of comparisons and those with $n\log n$ or $n\log^2 n$ comparisons, and in particular cannot imply that quicksort uses $O(n\log n)$ comparisons.

Let $D(n) = \log[C(n)/n]$, so that
$$
\lim_{n\to\infty} [D(2n) - D(n)] = 0.
$$
What we can conclude is that $D(n) = o(\log n)$, which translates to $C(n) = n 2^{o(\log n)}$.

Another problem with your analysis is that the number of comparisons could depend on the input, not only on the input size. In other words, $C(n)$ is not well-defined. It is known that quicksort has worst case number of comparisons $\Omega(n^2)$, but the average number of comparisons (over a random permutation) is only $\Theta(n\log n)$. We usually take $C(n)$ to be either the worst case number of comparisons or the average number of comparisons (for a random permutation).

Yet another problem is that it's not clear how to establish $C(2n)/C(n) \to 2$. Is this the result of an experiment, a heuristic calculation, or a formal proof? You don't make it clear. Experiments might show that $C(2n) \approx 2C(n)$ for small $n$, but the asymptotic behavior might be different; $C(2n)/C(n)$ might grow slowly, but eventually tend to infinity, for example.