The point here is the following result, that you can find in Zariski's book "Algebraic Surfaces", page 26. Zariski calls it "Extended Theorem of Bertini".

> **Theorem (Extended Bertini)**
> 
> **(1)** The general curve of an irreducible linear system cannot have multiple points outside the base locus of the system.
>
> **(2)** A reducible linear system, without fixed components, is necessarily composed with the curves of a pencil.

Here "reducible" [resp. "irreducible"] means that the general curve of the system is reducible [resp. irreducible].

Now, let us write $\mathcal{L}=Z+\mathcal{M}$, where $Z$ is the fixed part and $\mathcal{M}$ is the moving part. Then by Extended Bertini it follows that the general element $M \in \mathcal{M}$ is necessarily irreducible, **unless $\mathcal{M}$ is composed with a pencil**.

The last situation can occur. For instance let $S=\mathbb{P}^1 \times \mathbb{P}^1$, whose natural pencils are denoted by $|F_1|$ and $|F_2|$, and take $H=F_1+2F_2$ and $C \in |F_1|$. Then $H$ is very ample but $\mathcal{L}=|H-C|=|2F_2|$, which is without fixed part and composed with the pencil $|F_2|$. In fact, any element of $|2F_2|$ is the union of two curves in the pencil $|F_2|$, in particular it is **not** irreducible. 

**Remark.** The situation described in J. C. Ottem's comment is slightly different. In that example, indeed, we have a fixed part $Z=2E$; the moving part, however, is irreducible.