= Solution
Let $P$ now denote the degree-zero torsion <line bundle> represented by the point of $A[N]$, and put $L_P=\mathcal O_C(2K_C)\otimes P$. Again the <Riemann-Roch theorem> gives $h^0(L_P)=n$ and $h^1(L_P)=0$. For translation, use $\Theta_P=\Theta+P$ as a translated subset. With ordinary $T_P(x)=x+P$, this is $T_{-P}^*\Theta$; the notation $t_P^*\Theta=\Theta+P$ therefore uses $t_P=T_{-P}$.
Let $s_P$ and $s_0$ be the evaluation <determinants> for the two chosen bases, interpreted as sections of $\det E_{L_P}$ and $\det E_L$. The kernel for the first determinant is $H^0(C,2K_C+P-\sum z_j)$. By <Serre duality> and the <Riemann-Roch theorem> its dimension is $h^0(C,\sum z_j-K_C-P)$. Thus the determinant vanishes along $\alpha^*\Theta_P$, including multiplicities by the same determinant-complex argument as in part (i). The quotient is a rational section
$$
r=\frac{s_P}{s_0}\quad\text{of}\quad \det E_{L_P}\otimes(\det E_L)^{-1}\cong\bigotimes_{j=1}^n\operatorname{pr}_j^*P,
$$
with zero-minus-pole <Cartier divisor> $\alpha^*(\Theta_P-\Theta)$. The diagonal factors in the two determinants cancel.
Choose a trivialization $u:P^{\otimes N}\xrightarrow{\sim}\mathcal O_C$. This is possible because $P\in A[N]$. Taking the $N$th tensor power of $r$ and using $\bigotimes_j\operatorname{pr}_j^*u$ produces a scalar <rational function> $R$ on $C^n$ with
$$
\operatorname{div}(R)=N\alpha^*\Theta_P-N\alpha^*\Theta.
$$
On $X$, the <Theorem of the square> and $NP=0$ imply
$$
\mathcal O_X(\Theta_P-\Theta)^{\otimes N}=\lambda(-P)^{\otimes N}\cong\mathcal O_X.
$$
Consequently there is a <rational function> $f_P\in k(X)^\times$ with $\operatorname{div}(f_P)=N\Theta_P-N\Theta$. The map $\alpha$ is surjective, since every degree-$3g-3$ <line bundle> has a section by the <Riemann-Roch theorem>. The two <rational functions> $\alpha^*f_P$ and $R$ have the same <Cartier divisor> on the smooth projective connected product $C^n$; their quotient is an everywhere invertible regular function and hence a constant. Rescale $f_P$ to remove that constant. Away from the diagonals, ordinary evaluation gives
$$
\boxed{\alpha^*f_P=\left(\frac{\det(\sigma_i^P(z_j))}{\det(\sigma_i(z_j))}\right)^N,\qquad \operatorname{div}(f_P)=N\Theta_P-N\Theta,}
$$
where the $N$th power is interpreted with the chosen torsion trivialization. Equality as <rational functions> then extends from this dense open set to the whole product.
The <torsion trivialization in a determinant quotient> cannot be omitted if one uses raw meromorphic representatives. Explicitly, write $P=\mathcal O_C(p)$ and choose $h$ with $\operatorname{div}(h)=Np$. If $r_{\mathrm{raw}}$ is the quotient of the evaluation matrices in these meromorphic frames, then
$$
\operatorname{div}(r_{\mathrm{raw}})=\alpha^*(\Theta_P-\Theta)-\sum_j\operatorname{pr}_j^*p,\qquad R=r_{\mathrm{raw}}^N\prod_j h(z_j).
$$
Thus the displayed formula suppresses bundle trivializations; read as an unframed quotient of arbitrary meromorphic representatives, it needs precisely this extra product. Changes of bases or rescaling $u$ only change the allowable nonzero scalar normalization of $f_P$.
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