Francesco's answer is completely general.  As he emphasized, we should do everything in an intrinsic way so that there is a canonical choice of Theta.  I hope I get this straight since it is very pretty.

Indeed this is Riemann's classical proof of Jacobi inversion.  If L is any degree g line bundle, then K-L has degree g-2, so the translate C+(K-L) lies in Pic^(g-1). Then the pullback of O(Theta) = O(W(g-1)) by this translation map C-->Pic^(g-1)(C) is L.  (This also shows that this map induces an isomorphism from line bundles on Pic^(g-1)(C) of chern class [Theta], with line bundles on C of degree g.)

I.e. consider this as the problem of intersecting the curve translate C+(K-L) with the divisor W(g-1).  We know L = O(p1+...+pg) for some points pi, and that the intersection has degree g.  So if the intersection is proper, there exist g points p such that (p+ K -p1-...-pg) lies on W(g-1), i.e. such that

p + K-p1-...-pg = q1+...+qg-1, for some q's.

But this is true for p equal to any pi, i=1,....,g, if for p = pi0 we take 
the q's to be such that O(q1+...+qg-1) = O(K- sum of pi for i ≠ i0).  (Such q's exist by Riemann-Roch.)


Thus intersecting C+(K-L) with W(g-1) was Riemann's way of finding a divisor D such that O(D) = L.  Your question is the same, but phrased in terms of line bundles.  The common practice in many books of always working in Pic^0(C) via translating by a fixed point may obscure the intrinsic nature of the statements.  In fact I seem to be guilty of this in my argument on pp.386-7 of Lectures on Riemann Surfaces, (World Scientific Publishers 1989, ed. by Cornalba, Gomez-Mont and Verjovsky), where I take the strange approach of deducing this intrinsic version from a clunky non intrinsic version, thus making it look less comprehensible.

If you look closely at Lemma 3.4 of Lange Birkenhake, p.335, on which the exercise Francesco cites is based, you may discern a translated version of the Riemann result proved here.