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

Applying this to your situation, write $\mathcal{L}=Z+\mathcal{M}$, where $Z$ is the fixed part and $\mathcal{M}$ is the moving part. Then by the results above we see that the general element $M \in \mathcal{M}$ is necessarily irreducible, **unless $\mathcal{M}$ is composed with a pencil**.

This last situation can happen. For instance, let $S$ be a smooth quadric surface in $\mathbb{P}^3$ with rulings $|F_1|$ 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 composed with the pencil $|F_2|$. In fact, any element of $|2F_2|$ is the union of two lines in the ruling $|F_2|$, in particular is **not** irreducible.