Here are the answers to your questions.

1. The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$ To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection.

2. The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$ EDIT: Here is a justification. If $\pi : C^{d} \rightarrow C_{d}$ is the quotient map, then $\Delta$ is the branch divisor of $\pi.$ By relative duality we have $$({\pi}_{\ast}\mathcal{O}_{C^d})^{\vee} \cong {\pi}_{\ast}(K_{C^d} \otimes {\pi}^{\ast}{K}_{C_d}^{\vee}) \cong {\pi}_{\ast}\pi^{\ast}(\widetilde{K}_{C} \otimes K_{C_d}^{\vee}) \cong \widetilde{K}_{C} \otimes {K}_{C_d}^{\vee} \otimes {\pi}_{\ast}\mathcal{O}_{C^d}$$ We then must have that $\mathcal{O}_{C_d}(-{\Delta}/2) \cong \det{\pi_{\ast}\mathcal{O}_{C^d}} \cong \widetilde{K}_{C} \otimes K_{C_d}^{\vee}.$

Here are the answers to your questions.

1. The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$ To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection.

2. The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$ EDIT: Here is a justification. If $\pi : C^{d} \rightarrow C_{d}$ is the quotient map, then $\Delta$ is the branch divisor of $\pi.$ By relative duality we have $$({\pi}_{\ast}\mathcal{O}_{C^d})^{\vee} \cong {\pi}_{\ast}(K_{C^d} \otimes {\pi}^{\ast}{K}_{C_d}^{\vee}) \cong {\pi}_{\ast}\pi^{\ast}(\widetilde{K}_{C} \otimes K_{C_d}^{\vee}) \cong \widetilde{K}_{C} \otimes {K}_{C_d}^{\vee} \otimes {\pi}_{\ast}\mathcal{O}_{C^d}$$ We then must have that $\mathcal{O}_{C_d}({\Delta}/2) \cong \det(\pi_{\ast}\mathcal{O}_{C^d})^{\vee} \cong \widetilde{K}_{C} \otimes K_{C_d}^{\vee}.$

Here are the answers to your questions.

1. The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$ To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection.

2. The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$

Here are the answers to your questions.

1. The numerical class of the line bundle $\widetilde{L}$ is ${\rm deg}(L) \cdot x.$ To see why, note that the pullback of this class to $C^{d}$ must be ${\rm deg}(L)(X_{1}+ \cdots +X_{d}),$ where the divisor class $X_{i}$ on $C^{d}$ represents pullback of a point from the $i-$th projection.

2. The canonical bundle on $C_{d}$ is not $\widetilde{K}_{C},$ but rather $\widetilde{K_{C}}(-{\Delta}/2),$ where $\Delta$ is the diagonal divisor on $C_{d}.$ EDIT: Here is a justification. If $\pi : C^{d} \rightarrow C_{d}$ is the quotient map, then $\Delta$ is the branch divisor of $\pi.$ By relative duality we have $$({\pi}_{\ast}\mathcal{O}_{C^d})^{\vee} \cong {\pi}_{\ast}(K_{C^d} \otimes {\pi}^{\ast}{K}_{C_d}^{\vee}) \cong {\pi}_{\ast}\pi^{\ast}(\widetilde{K}_{C} \otimes K_{C_d}^{\vee}) \cong \widetilde{K}_{C} \otimes {K}_{C_d}^{\vee} \otimes {\pi}_{\ast}\mathcal{O}_{C^d}$$ We then must have that $\mathcal{O}_{C_d}(-{\Delta}/2) \cong \det{\pi_{\ast}\mathcal{O}_{C^d}} \cong \widetilde{K}_{C} \otimes K_{C_d}^{\vee}.$