Let $f:X\longrightarrow Y$ be a finite morphism of degree $d$ projective surfaces over $\mathbb{C}$. Suppose that $X$ is smooth. Let $L'$ be a line bundle on $Y$, and let $L=f^*L'$.
1) Consider a smooth curve $C\in |L|$. If the image $f(C)$ is a smooth curve on $Y$ which is a Cartier divisor, then does $f(C)\in |L'|$.
2) More generally if two curves $C_1$ and $C_2$ are linearly equivalent in $X$. Consider their images $C_1'$ and $C_2'$, and suppose that they Cartier divisors in $Y$. If $f|_{C}:C_i\longrightarrow C_i'$ are finite morphisms of degree $d$. Then are the images linearly equivalent too? I suppose that they are, because that linear equivalence is preserved by direct images.
So if $C_1 \sim C_2$, then $f_*(C_1)\sim f_*(C_2)$. That is $d.C_1'\sim d.C_2'$. Hence $C_1'\sim C_2'$. Is this right?
I see the flaw in my argument in part 2). If have $d.C_1'\sim d.C_2'$. This means that $\mathcal{O}(C_1')^d=\mathcal{O}(C_2')^d$. But this does not mean that $\mathcal{O}(C_1')=\mathcal{O}(C_2')$. For example we for a degree $d $ morphism, we have non trivial line bundle $L$ on $Y$ such that $L^d=\mathcal {O}_Y$