Yes, exactly and it holds in a more general setting under some added hypothesis:
If $X$ is an algebraic variety of dimension $n$ and $L$ is a line bundle over $X$ such that $h^0(L)=n+1$, $H^n\neq 0$, where $H$ is the mobile part of $L$, then the map $\phi_L: X \to \mathbb{P}(H^0(L)^*)\cong \mathbb{P}^{n} $ associated to $L$ is surjective.
A proof of the statement maybe can be the following one:
Let us suppose by contradiction that there is a point $p$ that is not in the image of the canonical map. Then you can choose $n$ hyperplanes of $\mathbb{P}^n$ whose common intersection is the point $p$. Then
$H_1.\cdots .H_n=0$ and so
$H^n=\phi_L^*(H_1).\cdots . \phi_L^*(H_n)=0$
that is a contradiction.
Thus the image of the map associated to $L$ has to be surjective.
In the case $n=1$ the hypothesis are always satisfied. In fact taking the mobile part $H$ of $L$, then $H\neq 0$ otherwise $h^0(L)<2$, and it is effective, by definition. This means $deg(H)>0$ and we have done.
Please note that if the map is a morphism then there isn't a base point in which any global section of $L$ vanishes on it, so that $L=H$. Pay attention that the viceversa does not hold.
Thus you get also a direct corollary:
If $h^0(L)=n+1$ and $L$ is base point free with $L^n\neq 0$, then the induced map $\phi_L$ is a surjective morphism.
Best Answer
If $g(C) = 2$ so that $C \subset J$ is a divisor, then by adjunction $$ K_C = (K_J + C)\vert_C = C\vert_C $$ since $K_J = 0$. So, indeed, $L = O_J(C)$ works.
Since the restriction map $Pic^0(J) \to Pic^0(C)$ is an isomorphism, this is the only line bundle with this property.