Let $X$ and $Y$ be smooth projective surfaces over an alg. closed field. Let $f:X\longrightarrow Y$ be a finite morphism of degree 2. Let $C$ be a smooth curve which maps to a smooth $C'$ under $f$. Is it obvious that $\mathcal{O}_X(C)=f^*\mathcal{O}_Y(C')$?
Conversely if $L'$ is a line bundle on $Y$, and if $C\in |f^*L'|$, then is $f(C)\in |L'|$?
a) No, it is not true that $\mathcal{O}_X(C)=f^*\mathcal{O}_Y(C')$.
For example take $X=\mathbb P^2_{x:y:z}, Y=\mathbb P^2_{u:v:w}, C=V(x)$ and let $$f(x:y:z)=(u:v:w)=(x^2:yz:z^2)$$ so that $C'= V(u)$.
Then $\mathcal O_X(C)=\mathcal O_{\mathbb P^2}(1)$, whereas $f^*(\mathcal O_Y(C'))=f^*(\mathcal O_{\mathbb P^2}(1))=\mathcal O_{\mathbb P^2}(2)$
b) The same set-up also gives a negative answer to your second question:
Take for $L'$ the line bundle $\mathcal O_{\mathbb P^2}(1)$. Then $f^*(L')=\mathcal O_{\mathbb P^2}(2)$ and you may take for $C$ the conic $z^2=xy$.
Its image $f(C)$ is the curve $C'$ given by $ w^3=uv^2$ which nor all thy Piety nor Wit will lure to $|L'|$ .