I am unable to follow the section of the solution I have underlined in green.
Let us revisit the calculation in your notes that shows that a shock can form in finite time starting from appropriate initial conditions that are piecewise continuous. The equation $$\frac{\partial u}{\partial t}+c(u)\frac{\partial u}{\partial x}=0,$$ and the initial condition $u_0(x)$ is constant and equal to $\alpha$ for $x\gt0$ and a smooth function for $x\lt0$ satisfying $\tfrac{\partial u}{\partial x}(0-,t)\lt0$.
$\quad$ Instead of using a curvilinear coordinate system near the dividing characteristic, calculate the jump in the derivative at $x=0$ as follows:
$\:\:(\rm i)$ Write down the equation for characteristics in the neighbourhood of $x=0$, i.e. for $\xi\lt0$.
$\,\,(\rm ii)$ Approximate the characteristics for small $\xi$ and show that they are given by $$x-c(\alpha)t=\xi\left[1+u^\prime_0(0-)c^\prime(\alpha)t\right]+{\cal O}(\xi^2).$$
$\,(\rm iii)$ Use this result to show that the jump in the derivative is given by $$\left[\frac{\partial u}{\partial x}\right]=\frac{u^\prime_0(0-)}{1+u^\prime_0(0-)c^\prime(\alpha)t},$$ $\quad$ and hence that blowup will occur in finite time.
$f(x+a)$ can be approximated as partial sums of the series $f(x) + af'(x) + {1 \over 2} a^2f''(x) + \ ...$. The approximation is progressively more accurate as more terms are taken.
The underlined section emphasizes that the remainder term, $...$, is small, so can be neglected. Hence an approximation for $f(x+\epsilon)$ when $\epsilon$ is small is $f(x) + \epsilon f'(x)$.