= Solution
Expand the two <Dirichlet polynomials>, keeping the <complex conjugate> on $B$. The diagonal $m=n$ contributes exactly $T\sum_{n\le X}a_n\overline{b_n}/n^{2\sigma}$. If $m\ne n$, the exponential integral is
$$
\int_T^{2T}e^{it\log(m/n)}\,dt=\frac{e^{2iT\log(m/n)}-e^{iT\log(m/n)}}{i\log(m/n)},
$$
whose absolute value is at most $2/|\log(m/n)|$. Thus \b[the mean-value formula] is
$$
\boxed{\int_T^{2T}A(\sigma+it)\overline{B(\sigma+it)}\,dt=T\sum_{n\le X}\frac{a_n\overline{b_n}}{n^{2\sigma}}+O\left(\sum_{\substack{m,n\le X\\m\ne n}}\frac{|a_n||b_m|}{n^\sigma m^\sigma|\log(m/n)|}\right).}
$$
This also covers $T=0$.
For $m>n$, integration of $1/x$ over $[n,m]$ gives $\log(m/n)\ge(m-n)/m$; the analogous inequality holds with $m,n$ exchanged. Hence $|\log(m/n)|^{-1}\le\max(m,n)/|m-n|\le X$. This gives the requested first bound $O\bigl(X\sum_{m\ne n}|a_n||b_m|/(n^\sigma m^\sigma)\bigr)$.
For the weighted bound, put $\alpha_n=|a_n|n^{-\sigma}$ and $\beta_m=|b_m|m^{-\sigma}$. The inequality
$$
2\alpha_n\beta_m\le\frac nm\alpha_n^2+\frac mn\beta_m^2
$$
reduces the estimate to a row sum and its symmetric counterpart. For a fixed $n$, the preceding logarithm bounds give
$$
\sum_{\substack{m\le X\\m\ne n}}\frac{n/m}{|\log(m/n)|}\le\sum_{m>n}\frac n{m-n}+\sum_{m<n}\frac{n^2}{m(n-m)}\ll n\log(2X).
$$
Here $n^2/(m(n-m))=n(1/m+1/(n-m))$, and each resulting <harmonic sum> is $O(\log(2X))$. The symmetric row sum is bounded by $O(m\log(2X))$. Therefore \b[the second error bound] is
$$
\boxed{O\left(\sum_{n\le X}\frac{|a_n|^2n\log(2X)}{n^{2\sigma}}+\sum_{m\le X}\frac{|b_m|^2m\log(2X)}{m^{2\sigma}}\right).}
$$
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