A perfect affine module sheaf admits a finite projective resolution

By [F1] choose a bounded finite-projective complex $Q$ whose associated sheaf complex represents $\tilde{M}[0]$. Full faithfulness reflects this equivalence to $Q\simeq M[0]$ in the module derived category. By [F3], $H^0(Q)\cong M$ and $H^i(Q)=0$ for $i\mathrm{\ne }0$. If the greatest nonzero degree $N$ is positive, vanishing of $H^N(Q)$ says $d^{N-1}:Q^{N-1}\to Q^N$ is onto. Lift the identity of $Q^N$ through this surjection by [F2], obtaining a section $s$. The kernel $K$ is a direct summand: the maps $Q^{N-1}\to K\oplus Q^N$, $u\mapsto (u-s{d}^{N-1}u,d^{N-1}u)$, and $(v,w)\mapsto v+s(w)$ are inverse. Hence $K$ is finite projective. The preceding differential lands in $K$ because consecutive differentials compose to zero. Replace $Q^{N-1}$ by $K$ and discard $Q^N$; the removed summand is the two-term identity complex $Q^N\to Q^N$, so cohomology is unchanged. Repeat finitely many times to obtain a bounded finite-projective complex $R$ concentrated in degrees at most zero, still with only $H^0(R)\cong M$. If there were no positive terms, take $R=Q$. The zero complex is allowed.

Since $R^1=0$, the quotient map $R^0\to H^0(R)\cong M$ is onto with kernel $\mathrm{im}{d}^{-1}$. For each negative degree, vanishing cohomology gives $\ker d^i=\mathrm{im}{d}^{i-1}$. At the leftmost nonzero degree $-d<0$ the preceding term is zero, so its differential is injective. Thus, putting $P_i=R^{-i}$ and adding zero terms if needed, the augmented sequence $0\to P_d\to \cdots \to P_0\to M\to 0$ is exact and finite, with finite projective terms. If $R$ is concentrated in degree zero this is the length-zero resolution; if $M=0$ the all-zero resolution is permitted. AC and the universe assumptions enter through [F1]; splitting each of the finitely many surjections uses only the projective lifting property. This proves the claim without additional ring hypotheses.