= Solution
Fix $p_0\in X$ and use the normalized <holomorphic one-forms> and <period lattice> of Solution 1. The <Abel-Jacobi map of a compact Riemann surface> is
$$
u(p)=\left(\int_{p_0}^p\omega_1,\ldots,\int_{p_0}^p\omega_g\right)\pmod\Lambda.
$$
Changing paths changes the vector by a period, so this is well defined and <holomorphic>. Adding these vectors gives a symmetric map on $X^d$ and therefore the <Abelian sum map> $u_d:X^{(d)}\to\operatorname{Jac}(X)$. Locally, even where points coincide, symmetric <holomorphic functions> descend to the elementary symmetric coordinates of the <symmetric product of a curve>. For any <divisor> $D=\sum_pn_pp$ of degree zero, define $u(D)=\sum_pn_pu(p)$. \b[<Abel's theorem> says that $D$ is a <principal divisor> if and only if $u(D)=0$.] Equivalently, two effective <divisors> of the same degree have the same Abel sum exactly when they are <linearly equivalent>.
We first establish the <residue> version of the bilinear identity. Let $\eta$ be a <meromorphic differential on a Riemann surface> with only simple poles and total <residue> zero, and let $F=\int_{p_0}^p\omega$ on the cut polygon, for a <holomorphic one-form> $\omega$. Excise small disks about the poles. Since $d(F\eta)=\omega\wedge\eta=0$ off the poles, the outer boundary integral is the sum of the positive small-circle integrals. The latter are $2\pi i\operatorname{res}_p(\eta)F(p)$. Pairing the outer edges as in Solution 1 therefore gives
$$
\sum_i\bigl(A_i(\omega)B_i(\eta)-B_i(\omega)A_i(\eta)\bigr)
=2\pi i\sum_p\operatorname{res}_p(\eta)\int_{p_0}^p\omega.
$$
The paths and the cut polygon are fixed together in this equality. Changing them changes the Abel vector by its <period lattice>, which is precisely the ambiguity we need.
For the sufficiency direction, we need a <differential of the third kind> with prescribed <residues> $n_p$. Let $S$ be the reduced support of $D$, consisting of $s$ points. When $s>0$, the <Riemann-Roch theorem> gives $h^0(K_X(S))=g-1+s$. The <residue> map from this space has <kernel> $H^0(K_X)$, of dimension $g$: a differential with at most a simple pole and <residue> zero has no pole. Its image therefore has dimension $s-1$. The <residue theorem> puts the image inside the $s-1$ dimensional space of tuples with sum zero, so it is all that space. In particular there is a <meromorphic differential on a Riemann surface> $\eta$ with <residues> $n_p$. Subtracting a combination of the normalized <holomorphic one-forms> makes all its $a$-periods zero. The bilinear identity then says
$$
(B_1(\eta),\ldots,B_g(\eta))=2\pi i\,u(D),
$$
where on the right we temporarily choose the representative determined by the paths. If $D=0$, we simply take $\eta=0$.
Suppose now that $u(D)=0$ in the <Jacobian variety>, so the chosen vector is $m+Bn$, with $m,n\in\mathbb Z^g$. Replace $\eta$ by
$$
\eta'=\eta-2\pi i\sum_jn_j\omega_j.
$$
Its $a$-periods are $-2\pi in_j$ and its $b$-periods are $2\pi im_i$. The periods around the punctures are $2\pi in_p$. These loops together generate the <homology> of the punctured <Riemann surface>, so every period of $\eta'$ lies in $2\pi i\mathbb Z$. Consequently $f(p)=\exp(\int^p\eta')$ is a single-valued nonzero <holomorphic function> off the support of $D$. At a support point, $\eta'=n_p\,dt/t+$ a <holomorphic> differential, so $f=t^{n_p}$ times a nonvanishing <holomorphic function>. It extends meromorphically and has <divisor> exactly $D$. This proves sufficiency, including negative coefficients.
Conversely, if $D=\operatorname{div}(f)$, take $\eta=df/f$. Its <residues> are the coefficients of $D$, while its $a$- and $b$-periods are integer multiples of $2\pi i$, by the winding numbers of $f$ along these loops. Writing them as $2\pi ia$ and $2\pi ib$, the <residue> bilinear identity gives $u(D)=b-Ba\in\Lambda$. Thus $u(D)=0$ in the <Jacobian variety>, proving necessity and hence
$$
\boxed{\operatorname{div}(f)=D\text{ for some meromorphic }f\ne0\ \Longleftrightarrow\ \deg D=0\text{ and }u(D)=0.}
$$
Here the displayed necessity of degree zero is also the <residue theorem> applied to $df/f$.
There is a useful geometric version of the conclusion. For an effective <divisor> $D$, the fiber of $u_d$ through $D$ consists precisely of the effective <divisors> <linearly equivalent> to $D$. A nonzero section of $\mathcal O_X(D)$ is a <meromorphic function> $f$ with $\operatorname{div}(f)+D\ge0$, and its zero <divisor> as a section is $\operatorname{div}(f)+D$. Two such sections give the same effective <divisor> exactly when their ratio is a nonzero constant. Thus the fiber is the <complete linear system of a divisor> $|D|=\mathbb P H^0(X,\mathcal O_X(D))$.
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