By definition we have
$$
\begin{align*}
u(\theta) & = \theta q(\theta)-t(\theta) \\
\\
u(\theta + \delta) & = (\theta + \delta) q(\theta + \delta)-t(\theta + \delta)
\end{align*}
$$
By incentive compatibility (where $\theta + \delta$ is the true type, $\theta$ is the false type) we have
$$
u(\theta + \delta) \geq (\theta + \delta) q(\theta)-t(\theta)
$$
Using these (I used square brackets for clearer notation, there is no mathematical function to them)
$$
\begin{align*}
u(\theta + \delta) - u(\theta) & =
\left[(\theta + \delta) q(\theta + \delta)-t(\theta + \delta)\right] -
\left[\theta q(\theta)-t(\theta)\right]  \\
\\
u(\theta + \delta) - u(\theta) & \geq
\left[(\theta + \delta) q(\theta)-t(\theta)\right] -
\left[\theta q(\theta)-t(\theta)\right]
\end{align*}
$$
This also holds if you divide by $\delta > 0$.

Similar argument for 2.8., using true type $\theta - \delta$ instead of $\theta + \delta$.