Edit :
After multiplication the inequality by $\Lambda (t)$ and then sum over $t\in \mathscr{T}$ we get , $\displaystyle \sum_{t\in \mathscr{T}}\frac{(\#B)^2}{u(t)}\Lambda(t)\le (\#B)\sum_{t\in \mathscr{T}}\Lambda (t)+\sum_{b,b'\in B\\b\not=b'}\sum_{t\in \mathscr{T}}\Lambda(t)\sum_{t|(b-b')}1$.
I take only the last term of the R.H.S. and rewrite it as: $\displaystyle \sum_{b,b'\in B\\b\not=b'}\sum_{t\in \mathscr{T}}\sum_{t|(b-b')}\Lambda(t)=\sum_{b,b'\in B\\b\not=b'}\sum_{t\in \mathscr{T}}\log(b-b')\le \log(X).\sum_{b,b'\in B\\b\not=b'}\sum_{t\in \mathscr{T}}1$.
How it comes $\log(2X).((\#B)^2-\#B)$ in the last line ?
$$\#B \leq u(t)^{\frac12} \left( \sum_{r \pmod t} Z(b,t,r)^2\right)^\frac12$$
Squaring and dividing by $u(t)$, we have
$$\frac{(\#B)^2}{u(t)} \leq \sum_{r \pmod t} Z(b,t,r)^2$$
\begin{align} Z(B,t,r)^2 &= \left( \sum_{b \in B, b \equiv r \pmod t} 1\right)^2 \\ &= \left( \sum_{b \in B, b \equiv r \pmod t} 1\right)\left( \sum_{b' \in B, b' \equiv r \pmod t} 1\right) \\ &= \sum_{b,b' \in B \\b,b' \equiv r \pmod t} 1\end{align} Edit to explain the next line: \begin{align}\sum_{r \pmod t} \sum_{b,b' \in B \\b,b' \equiv r \pmod t} 1 &= \sum_{r \pmod t} \left( \sum_{b \in B \\ b \equiv r \pmod t} 1+ \sum_{b,b' \in B , b \neq b' \\b,b' \equiv r \pmod t} 1 \right) \\ &= \sum_{r \pmod t} \sum_{b \in B \\ b \equiv r \pmod t} 1+\sum_{r \pmod t} \sum_{b,b' \in B , b \neq b' \\b,b' \equiv r \pmod t} 1 \\ &= \sum_{r \pmod t} Z(B,t,r)+ \sum_{b,b' \in B , b \neq b' \\b' \equiv b \pmod t} 1 \\ &= \#B + \sum_{b , b' \in B \\ b \neq b' } \sum_{t|b-b'}1\end{align}
Edit to answer the edit:
\begin{align}\displaystyle \sum_{b,b'\in B\\b\not=b'}\sum_{t\in \mathscr{T}}\sum_{t|(b-b')}\Lambda(t)&=\sum_{b,b'\in B\\b\not=b'}\log(|b-b'|)\\&\le \log(2X).\sum_{b,b'\in B\\b\not=b'}1 \\ &= \log(2X) ((\#B)^2 - \#B) \end{align}