Generally, the natural density of a set $A\subseteq \{1, \dots, n\}$, denoted by $d(A)$, is defined as $$d(A)= \lim_{n\to\infty} \frac{\mid A \mid }{n}$$
Now, I am wondering, is there a similar notion over a bounded set $B\subset\mathbb{Z}$ instead of the set $\{1, \dots, n\}$. For instance, $B=\{1\leq i\leq n: i\equiv a\mod q\}$, and we are interested in the density of the set $A\subseteq B$ where $A= \{j\in B: m\mid j\}$ with $m\leq n $.
A more concrete instance can be computing the density of multiple of number $m$ in the set of odd numbers.
Hope this is clear.
I think you mean for the natural density to be $$d(A) = \lim_{n \to \infty} {|A \cap [n]| \over n}$$, where $[n] = \{ 1, 2, \ldots, n \}$.
In that notation, $d_B(A)$, the density of A in B, could be defined as
$$d_B(A) = \lim_{n \to \infty} {|(A \cap B) \cap [n]| \over |B \cap [n]|}$$.
This seems like a natural definition and can be found in a few places - for example the OEIS wiki mentions it, as does this paper of Jha and Sanna in the special case where B is the primes, this 2016 MathOverflow question, this 2023 blog post by John Cook... I'm having trouble finding a formal definition of "relative density" but I think this would be well-understood.
Then to your concrete example: let $B$ be the even numbers and let $A$ be the multiples of three. Then $(A \cap B) \cap [n]$ is the set of multiples of 6 less than $n$, so you have
$$d_B(A) = \lim_{n \to \infty} {\lfloor n/6 \rfloor \over \lfloor n/2 \rfloor} = {1 \over 3}$$
But if we take $A$ to be multiples of 4, then $(A \cap B) \cap [n]$ is the set of multiples of 4 less than $n$, and you have
$$d_B(A) = \lim_{n \to \infty} {\lfloor n/4 \rfloor \over \lfloor n/2 \rfloor} = {1 \over 2}.$$
Of course $B$ has a natural density here so this is just the quotient of natural densities. But consider for example $B$ the primes and $A$ the primes of the form $4n+1$; then it's well-known that the density of $A$ in $B$ is 1/2, though $A$ and $B$ both have natural density 0.