In general, the answer is yes. Let's assume that $C$ is geometrically irreducible curve over any field $k$.
Note that we have a short exact sequence of sheaves
$$1\to \mathcal{O}_C^\times\to v_\ast \mathcal{O}_{C'}^\times\to (v_\ast\mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times\to 1$$
Taking the long exact sequence in cohomology gives
$$1\to k^\times\to k^\times\to ((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)(C)\to \mathrm{Pic}(C)\to \mathrm{Pic}(C')\to 1$$
where we note that $(v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times$ is finitely supported which is why its cohomology is zero. In particular, if $C$ has $n$ double points then it's not hard to see that
$$((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)(C)=\bigoplus_{x_i\text{ node}}((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)_{x_i}=(k^\times)^n$$
essentially because locally around a node $x_i$, with preimage points $p_1$ and $p_2$, we have that the curve has ring of functions $\{f(t)\in k[t]:f(p_1)=f(p_2)\}$ and locally around each of the two points $p_1$ and $p_2$ in $v^{-1}(x_i)$ we have that the functions just look $k[t]$ so the isomorphism $((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)_{x_i}\to k^\times$ is something like $f\mapsto f(p_1)f(p_2)^{-1}$.
All in all we see that we have a short exact sequence
$$1\to (k^\times)^n\to\mathrm{Pic}(C)\to\mathrm{Pic}(C')\to 1$$
so that $\mathrm{Pic}(C)\to\mathrm{Pic}(C')$ is essentially never injective.
This is just a record of Aphelli's suggestion, as an answer.
Write $ P $ as $ \mathbb{P} (E) $ for a holomorphic rank $ 2 $ vector bundle $ E $ on $ X $. We will use the following theorem of Serre.
Theorem: Let $ X $ be a projective variety and $ E $ a vector bundle on $ X $. Let $ H $ be a very ample line bundle on $ X $. Then there is an integer $ N $ (which depends on $ H $ and $ E $ in general) such that for every $ n \ge N $, the vector bundle $ E \otimes H^{\otimes n } $ is generated by its global sections - For clarification, this means that the global sections restricted to a fiber span the fiber as a vector space.
Remark: The space of global sections of $ E \otimes H^{\otimes n} $ is finite dimensional and for large enough $ n $ as above, this dimension grows asymptotically at the rate of $ n^{\dim X} $. (This is not strictly needed but good to know.)
Noting that a compact Riemann surface is a projective variety, we pick a very ample line bundle $ H $ on $ \Sigma $. Now choose an $ N $ for the pair $ E,H $ according to the theorem. Then $ E \otimes H^{\otimes N} $ has enough global sections. A general global section of the above vector bundle does not vanish on $ \Sigma $ as it is of rank $ 2 $ while $ \Sigma $ has dimension $ 1 $. (For a rigourous explanation of what 'general' means, I recommend reading about Bertini's Lemma)
Pick such a general section $ s $ as above. Note that the fibers of $ \mathbb{P}(E \otimes H^{\otimes N}) \rightarrow \Sigma $ are just projectivizations of the fibers of $ E \otimes H^{\otimes N} $, more or less, by definition. Since $ s $ does not vanish in any fiber, its value in every fiber's projectivization defines a section $ [s] $ of $ \mathbb{P}(E \otimes H^{\otimes N}) \rightarrow \Sigma $.
Finally $ \mathbb{P}(E \otimes H^{\otimes N}) $ is just $ \mathbb{P}(E) = P $ by the remark that tensoring by any line bundle does not change the projective bundle, hence $ [s] $ indeed is a section of $ P \rightarrow \Sigma $ as desired.
Best Answer
This is basically the content of Hartshorne exercise III.12.4:
The key result here is (a corollary to) the semicontinuity theorem:
So applying this to our situation at hand, we may see that the pushforwards of your line bundle trivial on the fibers is again a line bundle on the target. After some messing about with pushforwards and pullbacks (see here, for instance), one may see that your line bundle trivial on the fibers is indeed the pullback of a line bundle on $Y$.