I am not understanding the parts marked with "?" of this example. I understand what he is trying to show, which is that Theorem $6$ may not be valid in an infinite dimension space, and for this it will exhibit a linear functional, in this case $L (f)$ such that there is no $g$ satisfying $L (f ) = (f | g)$. But the marked steps really do not understand. Can anyone explain?
Remember we can consider the polynomial $h=x-z$ as a function $x\mapsto h(x)=x-z$, and so we can think of the polynomial $hf$ as the function $x\mapsto h(x)f(x)$. For your first question mark we have $hf(x)=(x-z)(f(x))$ for any polynomial $f$, so $hf(z)=0$ as claimed. Now by definition of $L$ we have $L(hf)=hf(z)=0$, but by our assumption (remembering that $hf$ is still a polynomial) we have the existence of some polynomial $g$ such that $$0=L(hf)=\int_0^1h(t)f(t)\overline{g(t)}dt.$$