Let me try to give a slightly different approach, which hopefully complements Jason's answer. Basically, let us compute as much as we can about the natural line bundles on the symmetric power.

First of all, the line bundle $O(-\Theta)$ on $\mathrm{Pic}^{g-1}(C)$ coincides with the line bundle $\det \mathrm{R\Gamma}$ (whose fiber over $\ell\in\mathrm{Pic}^{g-1}(C)$ is $\det\mathrm{R\Gamma}(C,\ell)$). I prefer to work with degree $g-1$ bundles on $C$ here, so that the $\Theta$ divisor is defined canonically, not just up to a shift. Your formula involves the pullback of $\Theta$ (or, equivalently, of this line bundle) under the map $$u_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(D+(-d+g-1)p_0),$$ but I would prefer to work with the map $$v_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(-D+(d+g-1)p_0),$$ which is algebraically equivalent to $-u_d$ (and $\Theta$ is invariant under the inversion anyway).

Now in these terms, we can state the following identity: $$v_d^*(\det\mathrm{R\Gamma}^{\otimes 2})\simeq O(\Delta-2(d+g-1)H),$$ where following Jason's answer, $\Delta$ is the diagonal in $\mathrm{Sym}^d C$, and $H$ is the reduced divisor whose complement is $\mathrm{Sym}^d(C-{p_0})$. (Note that this is an isomorphism of line bundles, not just an algebraic equivalence.) Indeed, the left-hand side is the line bundle whose fiber over $D\in\mathrm{Sym}^dC$ is $$\det\mathrm{R\Gamma}(C,O(-D+(d+g-1)p_0))^{\otimes 2}\simeq\det(O/O(-D))^{\otimes-2}\otimes (O(-D))_{p_0}^{\otimes 2(d+g-1)}.$$ It now remains to notice that the line bundle whose fiber over $D$ is $\det(O/O(-D))^{\otimes 2}$ is isomorphic to $O(-\Delta)$, while the line bundle whose fiber is $O(-D)_{p_0}$ is isomorphic to $O(-H)$.