I'll stick to the pullbacks and line bundle sides of things, since this is what you seem to be concentrating on.
Your first two bits are correct. If $f:X\to Y$ is a map of schemes, there is always a homomorphism $\renewcommand{\Pic}{\operatorname{Pic}}\renewcommand{\ClGrp}{\operatorname{Cl}} f^*:\Pic Y \to \Pic X$ given by sending $L\in\Pic Y$ to $f^*L$, a line bundle on $X$. (Though I don't see why you appeal to the fiber product to define $f^*$: the definition is just $f^*(-)=\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_Y} f^{-1}(-)$.)
When $f$ is only a rational map, things may break in interesting ways. If you want a map on Picard groups, then what you might think of doing is pulling back a line bundle $L$ on $Y$ to $U\subset X$ which is a domain of definition for the rational map $f$ and extending this to a line bundle on $X$. But there are obstacles: this line bundle may not extend uniquely (take $X=\Bbb P^1$, $Y=\Bbb A^1$, and $f$ the identity on $U_0$: then $f^*\mathcal{O}_Y$ extends to any line bundle on $X$), or the unique extension of $f^*L$ may not be a line bundle.
If you want to salvage a map of Picard groups here, you need to guarantee that any line bundle on the maximal domain of definition of $f$ has a unique extension. The best general result in this area is that a vector bundle on an open subscheme $U\subset X$ extends uniquely to a vector bundle on $X$ when $X$ is $S_2$ and $\operatorname{codim}(X\setminus U)\geq 2$ (ref). In particular, since you always deal with projective normal varieties over $\Bbb C$, your varieties are $S_2$ by Serre's criterion for normality and the indeterminacy locus of your morphism is at least codimenstion two by Hartshorne lemma V.5.1, for instance. So everything's okay.
Let $f \colon X \to Y$ be am etale double covering of a general plane sextic curve and let $L$ be the restriction of $\mathcal{O}(1)$ from the plane. Then
$$
H^0(X, f^*L) \cong H^0(Y, f_*f^*L) \cong H^0(Y, L) \oplus H^0(Y, L \otimes \xi),
$$
where $\xi$ is a line bundle of order 2 (i.e., such that $\xi^2 \cong \mathcal{O}_Y$) corresponding to the double covering. So, it is enough to check that $H^0(Y, L \otimes \xi) = 0$.
First, note that if $D$ is an effective divisor corresponding to a global section of $L \otimes \xi$ then $2D$ corresponds to $L^2$, hence it is cut out on $Y$ by a conic on the plane. Since it has the form $2D$, the conic is everywhere tangent to $Y$. Thus, it is enough to check that a general plane sextic does not have everywhere tangent conics.
Consider the variety $M$ of triples $(Y,C,D)$, where $Y$ is a smooth plane sextic, $C$ is a conic and $D$ is a divisor (of degree 6) in $Y$ such that $Y \cap C = 2D$. Note that $D$ is also a divisor on $C$. Moreover, the space $N$ of pairs $(C,D)$ has dimension $5 + 6 = 11$. Moreover, since plane sextics cut out a complete linear system (of dimension 12) on $C$, it follows that the fiber of the projection $M \to N$ has codimension 12 in the space of all sextic curves. Therefore, the dimension of $M$ is less by 1 than the dimension of the space of all plane sextics, hence the projection of $M$ to this space is not dominant, hence a general $Y$ is not in the image.
Best Answer
The assumption that $|D|$ is base-point free means that there is a surjective morphism $$ V \otimes \mathcal{O}_X \to \mathcal{O}_X(D). $$ The pullback functor is right-exact, hence the pullback of the above morphism $$ V \otimes \mathcal{O}_Y \to \pi^*\mathcal{O}_X(D) = \mathcal{O}_Y(\pi^*D) $$ is surjective, which means that $\pi^*D$ is base-point free.