In the (wonderful) book by C. Birkenhake and H. Lange Complex Abelian Varieties we can find the following result, see Corollary 4.3.4 page 77. It is stated in any dimension $g \geq 2$, but let us consider only the case of abelian surfaces for the sake of simplicity.
Proposition. Let $f \colon X \to Y$ be an isogeny of abelian surfaces and let $D$ be a positive definite and irreducible divisor on $Y$. Then $f^*D$ is also irreducible.
The proof starts as follows:
Assume the contrary, then $f^*D$ is a sum of effective divisors $D_1+ \cdots + D_n$. But necessarily $D_i \cdot D_j=0$ and $D_i$ is numerically equivalent to $D_j$ for all $i \neq j$, the map $f$ being an étale Galois covering $\ldots$
and then a contradiction is reached by using the Nakai-Moishezon theorem.
Now, I do not understand the part $D_i \cdot D_j=0$ for all $i \neq j$.
If $D$ is smooth then this is immediate: in fact, an étale cover of a smooth curve must be smooth, in particular its irreducible components do not intersect.
However, if $D$ is singular (and in the Proposition there is no smoothness assumption) the same argument do not work. Indeed, it is well-known that there are irreducible, $1$-nodal curves with a connected, étale double covering consisting of two copies of the normalization, see for instance example 10.6 in Chapter III of Hartshorne's Algebraic Geometry. The two components intersect at two points, that are the two nodes of the covering.
So my question is
Q. Is the result stated in the Proposition above also true when $D$ is singular? If not, what is a counterexample? Or maybe am I missing something?