Why Djikstra's algorithm is said to have $\mathcal{O}(|V|^2)$ complexity?
Algorithms are usually analyzed under the word RAM model, in which basic operations on machine words cost $O(1)$. Machine words are defined as words of length $\log n$, where $n$ is the size of the input (in bits).

In the case of algorithms on weighted graphs, we could make the further assumption that weights are put in special registers for which "reasonable" arithmetic operations are $O(1)$. This assumption is made implicitly, without any mention of it, since the exact computation model isn't really needed when designing an algorithm, as long as you're not "cheating". It allows using floating point weights, for example.

The computation model serves two purposes:


To explain what constitutes "cheating", that is, what is not allowed (or rather, what is allow).
To allow proving lower bounds.


If we are not careful with the computation model, we could solve PSPACE-complete problems such as TQBF; see here.

One instantiation of the murky class of computation models for weighted graphs is allowing the weights to fit in a constant number of machine words; then we can just analyze the algorithm using the word RAM model. In this case, any weight encountered during Dijkstra's algorithm also fits in a constant number of machine words, since any such weight is the sum of at most $|V|-1$ weights. Therefore in the word RAM model, assignment (as well as addition) costs $O(1)$ rather than $O(\log s)$.