I will assume that $n>j$ as you seem to be saying this in your question.
I'm afraid that you're approach won't work, as $m(I\cap J)\neq m(I)m(J)$ for the individual intervals $I,J$, with a simple counterexample provided by $I = J = [0,1/2)$. The way I would approach this problem would be to figure out how many intervals $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ are contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$ and add up these measures, then figure out the measures of the at most two intervals of the form $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\cap \left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$ which are nonempty but not the entire first interval. I will denote their measures $m_1,m_2$.
The first of these two intervals corresponds to a value of $r$ such that $\frac{k+r2^i}{2^{i+n}}\leq \frac{l}{2^j} < \frac{k+1+r2^i}{2^{i+n}}$ which we successively rearrange as $k+r2^i\leq \frac{l2^{i+n}}{2^j} < k+1+r2^i$ $r\leq l2^{n-j} - k2^{-i} < r+2^{-i}$ which can never be satisfied. To see this, note that $n>j\implies l2^{n-j}\in \mathbb{N}$, so we would need $x - 2^{-i}< k2^{-i}\leq x$ for some $x\in \mathbb{N}$, yet this means $x2^i - 1 < k \leq x2^i$ and so since $k\in \mathbb{N}$, $k = x2^i$, violating $k < 2^i$. Thus $m_1 = 0$ as the intersection of the intervals is empty.
The second of these two intervals corresponds to a value of $r$ such that $\frac{k+r2^i}{2^{i+n}}< \frac{l+1}{2^j} \leq \frac{k+1+r2^i}{2^{i+n}}$ which we successively rearrange as $k+r2^i< \frac{(l+1)2^{i+n}}{2^j} \leq k+1+r2^i$ $r< (l+1)2^{n-j} - k2^{-i} \leq r+2^{-i}$ which can never be similarly never be satisfied, so $m_2 = 0$ as well.
Al that remains is to count the number of intervals $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$. Each such interval corresponds to a value of $r$ satisfying $\frac{l}{2^j}\leq \frac{k+r2^i}{2^{i+n}} < \frac{k+1+r2^i}{2^{i+n}} < \frac{l+1}{2^j}$ and since we already showed no interval $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ intersects but is not contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$, we can simplify this to $\frac{l}{2^j}\leq \frac{k+r2^i}{2^{i+n}} < \frac{l+1}{2^j}$ $\frac{l}{2^{j-i-n}}\leq k+r2^i < \frac{l+1}{2^{j-i-n}}$ $l{2^{n-j}} - k2^{-i}\leq r < (l+1)2^{n-j} - k2^{-i}$ and since $0 < k2^{-i} < 1$ we have exactly $2^{n-j}$ possible values for $r$. Furthermore, all of these possible values for $r$ are actually iterated over in $A$, as $n>j,l < 2^j\implies (l+1){2^{n-j}} - k2^{-i} < 2^n$ so $r\leq 2^n - 1$ as at most the greatest integer less than $(l+1){2^{n-j}} - k2^{-i}$. Thus our final answer is $m(A\cap B) = 2^{n-j}m\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\right) = 2^{n-j}\frac{1}{2^{i+n}} = \frac{1}{2^{i+j}}$
EDIT: The measure of $A$ is $2^nm\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\right) = 2^n\frac{1}{2^{i+n}} = \frac{1}{2^i}$, while the measure of $B$ is clearly $\frac{1}{2^j}$, so we do in fact have $m(A\cap B) = m(A)m(B)$.