[I have edited this below to hopefully clarify the topic and to specifically answer the question asked.]
As Francesco 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 $\mathrm{Pic}^{g-1}$. Then the pullback of $\mathscr O(\Theta) = \mathscr O(W(g-1))$ by this translation map $C\to\mathrm{Pic}^{g-1}(C)$ is $L$. (This also shows that this map induces an isomorphism from line bundles on $\mathrm{Pic}^{g-1}(C)$ 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 $L =\mathscr O(p_1+...+p_g)$ for some points $p_i$, and that the intersection has degree $g$. So if the intersection is proper, there exist $g$ points $p$ such that $(p+ K -p_1-...-p_g)$ lies on $W(g-1)$, i.e. such that
$p + K-p_1-...-p_g = q_1+...+q_{g-1}$, for some $q$'s.
But this is true for $p$ equal to any $p_i$, $i=1$,....,$g$, if for $p = p_{i_0}$ we take
the $q$'s to be such that $\mathscr O(q_1+...+q_{g-1}) = \mathscr O(K- \sum_{i\ne i_0}p_i)$. (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 $\mathrm{Pic}^0(C)$ 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.
Let me enlarge a little on this. Classically the theta divisor is the zero locus of a certain holomorphic theta function. Computing its zeroes is done by the residue theorem. Recall there is a generalization of that theorem that allows one to compute not just the number of zeroes inside a given loop but also the sum of their values, (see Ahlfors,4.5.2 the argument principle, pp. 151-153, esp. eq.(47) p.153).
Your question is exactly this, i.e. not only how many intersection points exist, but what is their sum as a point of the Jacobian, i.e. as a line bundle. Traditionally this was a lemma used to prove Riemann's theorem that the zeroes of the theta divisor are, up to a precise translation, exactly the image of the Abel map on $C^{g-1}$. This is made very clear in Griffiths-Harris, where the number of zeroes is computed on pages 334-335, and their sum on pages 336-337, both by the residue theorem. These results are then applied to prove Riemann's theorem on pages 338-339.
This result appears in the classical form also in Mumford's Tata lectures on theta I p.149, or Siegel's Topics on complex function theory vol. 2, section 4.10. If one uses a modern approach to that theorem by equating the homology classes of those two divisors and proving this forces them to be translates, as in Lange and Birkenhake, one loses some explicitness of the classical approach. It does however then make available the relatively easy intrinsic proof given here for your question, and resembling that of their lemma 3.4 on p.335 of their book. I think in some sense this may be a more natural approach.
Note that this result is key to the construction of a "Poincare" bundle, since it gives a uniform way to construct a family of divisors on $\mathrm{Pic}^g(C)$, i.e. a line bundle on $C\times\mathrm{Pic}^g(C)$, whose restrictions to the curve $C$ induce all line bundles of degree $g$. Hence this construction too is essentially due to Riemann. This is clearly explained by George Kempf on pages 154-157, chapter 18, of his book Abelian Integrals, Monografias del instituto de matematicas, #13, Universidad Nacional Autonoma de Mexico, 1983.
**edit**: Ten years on, I notice that neither response provided here actually states the answer to the original question posed. Perhaps that is why neither has been accepted. The OP asked for a specific intersection divisor; the first answer computes a different specific intersection, while this answer gives a general intersection formula, but does not use it to answer the OP’s question.
I wish to make three points: first, the OP’s example (as well as the example in the other answer), is easily derived from the general formula above; second, the truth of that general formula is almost trivial; third, the more trivial point of view leads directly to a construction of the universal Poincare divisor. (I will compute with intersection divisors, to deduce the result for pull backs of line bundles.)
In more detail:
1) Although there is no intrinsic theta divisor in Pic(0), there is an intrinsic one in Pic(g-1), namely W(g-1), and all other theta divisors are translates of this one. If C denotes the intrinsic copy of the curve in Pic(1), we are interested in intersecting translates of C with translates of W(g-1). Since intersection divisors are invariant under translation, this can always be equated to an intersection, in Pic(g-1), of W(g-1) with a translate of C.
E.g. if D,E are divisors, and we want to intersect the translate C+D with the translate W(g-1)+E, then 1 + deg(D) = (g-1) + deg(E), so deg(D-E) = g-2, and the desired intersection equals the intersection of W(g-1) with C + (D-E). By the general formula above, W(g-1).(C+L) = K-L, so the present intersection (C+D).(W(g-1)+E) = K-(D-E). [This is true as divisors in case the intersection is proper, and as line bundles otherwise. In particular, if D-E is itself effective, then p+(D-E) is effective for every choice of p, so the translate C+(D-E) lies inside W(g-1) hence C+D lies inside W(g-1)+E.]
In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general should not be effective, so the intersection divisor = g.p in that case, and in every case the pulled back line bundle is O(gp).
2) The formula W(g-1).(C+(K-L)) = L, explained above, has another simpler explanation. Note there is another intrinsic copy of the curve in the Picard variety, namely -C in Pic(-1). Translates of this curve are even easier to intersect with W(g-1). Specifically, if D is a general effective divisor of degree g, then the translate D-C has degree g-1, and W(g-1).(D-C) = D. This is essentially trivial, i.e. given D = p1+…+pg, which p satisfy D-p is effective? Certainly, this holds if p is one of the points p1,…,pg, and if D is general, i.e. if h^0(D) = 1, these are the only possibilities.
If we use the fact that W(g-1) is invariant under the involution taking L to K-L, (which uses Riemann-Roch), it follows that W(g-1) also intersects K-(D-C) = C + (K-D) in the divisor D.
3) Now consider the map C x Pic(g) —> Pic(g-1) taking (p,L) to L-p, and use it to pull back the intrinsic theta divisor W(g-1). If L = O(p1+…+pg) = O(D) is general, the fact that (D-C).W(g-1) = D, says that the restriction to C x {L} of the pull back of W(g-1) is precisely D. Thus as line bundles, the restriction to C x {L} of the pull back of O(W(g-1)) is L. I.e. the pull back of O(W(g-1)) is the "Poincare" line bundle of degree g. If E is any divisor of degree n-g, translating the Poincare divisor on CxPic(g) by the map CxPic(g)—>CxPic(n) sending (p,L) to (p,L+O(E)), gives a line bundle whose restriction to Cx{L+O(E)} is L, so adding on the divisor ExPic(n) gives a Poincare divisor of degree n.