The rationality theorem in birational geometry I am now reading the proof of rationality theorem in birational geometry of algebraic varieties written by Kollár-Mori. (pp.86)The main confusing  thing is the first step which reduced the big and nef divisor $H$ to the base point free case.
Since $H$ is big and nef, by using the property, we can write it linearly equivalent to the sum of a $\mathbb{Q}$-Cartier ample divisor $A_k$ and $\frac{1}{k}E$ for any sufficiently large $k$ and some fixed effective divisor $E$, but how can it be still nef when change it into a linear combination of $H$ and $K_X+\Delta$? I see it is still big by change the coefficient, but how can it be nef? 
Here being nef is more important since we want to use the base point free theorem proved in the former section.
Any help and hints are appreciated.
 A: Your concerns are right. They need to take an appropriate linear combination (not any!) of $K_X+\Delta$ and $H$. The way to do it is to appeal to the basepoint-free theorem (if you look at the proof you referred to, they say that $H'$ is basepoint-free by (3.3), which is the basepoint-free theorem).
The second observation is that we can assume that $r(H)>0$. Indeed, the theorem would hold right away in this case. This is not explicitly mentioned in the proof, and it may be what confused you. Now, with this said, we can go over the strategy more clearly:
We start with $H$ nef and big, and we want to show that $r(H)$ is a rational number with suitable properties. Since we may assume that $r(H)>0$, for $n \in \mathbb N$ very large, we have that $H+\frac 1 n (K_X + \Delta)$ is nef. Now, let $a$ as in the statement be such that $a(K_X+\Delta)$ is Cartier. Clearing denominators, we can rephrase everything saying that $naH + a(K_X+\Delta)$ is nef Cartier, and that 
$$
nH= (nH + (K_X+\Delta))-(K_X+\Delta)
$$
is nef and big. But this fits into the hypotheses of the basepoint-free theorem. Therefore, $|b(naH+a(K_X+\Delta))|$ is basepoint-free for $b$ large enough.
Then, you can reconcile to the exposition in the book putting $d=1$, $m=b$, and $c=na$.
