(Making my comment into an answer.)
The pullback of a vector bundle is always a vector bundle. The pushforward of a nontrivial vector bundle by a nontrivial embedding is never a vector bundle — it is trivial outside the image of the embedding.
Let $f\colon E\to X$ be any schematically dominant morphism of schemes, e.g., a surjective morphism with $X$ reduced. Let $U_{E/X}$ be the cokernel of the canonical morphism $f^{\flat}\colon\mathbb G_{m,X}\to f_{*}\mathbb G_{m,E}$ of etale sheaves on $X$ (this morphism is injective by the schematically dominant hypothesis. The sheaf just defined is, or should be, called the sheaf of relative units of $E$ over $X$). Then there exists a canonical exact sequence of abelian groups
$$
0\to\varGamma(X, \mathbb G_{m,X})\to\varGamma(E, \mathbb G_{m,E})\to
U_{E/X}(X)\to\textrm{Pic}(X)\overset{f^{*}}{\to}\textrm{Pic}(E).
$$
So, you must prove that in your particular case the map $\varGamma(E, \mathbb G_{m,E})\to U_{E/X}(X)$ is surjective. Well, is it?
Regarding the proof: exactness of the above exact sequence boils down to checking that the kernel of $f^{*}\colon \textrm{Pic}(X)\to\textrm{Pic}(E)$ is the same as the kernel of $H^{1}(f^{\flat})$. This follows from the fact that $f^{*}$ factors as
$$
H^{1}(X,\mathbb G_{m,X})\to H^{1}(X,f_{*}\mathbb G_{m,E})\hookrightarrow H^{1}(E,\mathbb G_{m,E}),
$$
where the first map is $H^{1}(f^{\flat})$ and the second map is the first edge morphism in the Cartan-Leray spectral sequence $H^{ r}(X, R^{ s} f_{*}\mathbb G_{m,E})\Rightarrow H^{ r+s}(E, \mathbb G_{m,E})$.
Observe now the following helpful fact: If $f$ has a section $\sigma\colon X\to E$, then $\mathbb G_{m,E}\to \sigma_{*}\mathbb G_{m,X}$ induces a morphism $f_{*}\mathbb G_{m,E}\to f_{*}\sigma_{*}\mathbb G_{m,X}=\mathbb G_{m,X}$ which splits the sequence
$$
0\to \mathbb G_{m,X}\to f_{*}\mathbb G_{m,E}\to U_{E/X}\to 0.
$$
In this case the surjectivity of $\varGamma(E, \mathbb G_{m,E})\to U_{E/X}(X)$ is evident. Does your map have a section?
Best Answer
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.