I guess Voisin means that if $X$ is projective then every divisor $Z$ homologous to $0$ is *linearly equivalent* to a divisor $Z'$ which is a sum of divisors with multiplicity $1$.

In fact, the element $\alpha_Z \in \textrm{Pic}^0(X)$ she wants to define only depends on the linear equivalence class of $Z$.

We start by recalling that if $X$ is projective then every divisor $Z$ is linearly equivalent to $Z_1 -Z_2$, where $Z_i$ is very ample. In fact, choose a very ample divisor $H$ and $m >>0$ such that $mH+Z$ is very ample, and set
$$Z_1:=mH+Z, \quad Z_2:=mH.$$
Now, since $Z_1$ and $Z_2$ are very ample, it is possible to choose $Z_1'$ linearly equivalent to $Z_1$ and $Z_2'$ linearly equivalent to $Z_2$ such that

- every irreducible component of $Z_1'$ and $Z_2'$ appears with multiplicity $1$;
- no component of $Z_2'$ is contained in $\textrm{Supp}(Z_1')$.

Therefore $Z':=Z_1'-Z_2'$ is a divisor linearly equivalent to $Z$ and with the desired property.

By the way, I do not think that the assumption "degree $0$" is really important here.

linearly equivalentto a divisor in which every irred. hyp. appears with multiplicity 1. $\endgroup$ – Francesco Polizzi Jul 14 '11 at 8:09