Hypercover total degree uses only a finite matching truncation

Degree bookkeeping. In the total complex of the hypercover, the term in bidegree $(p,q)$ has $p\ge 0$ and $q\ge a$ and contributes to total degree $p+q$. For a fixed total degree $m$ this gives $0\le p\le m-a$. The differentials entering and leaving total degree $m$ involve terms of total degree $m-1$, $m$ and $m+1$, so the largest simplicial level appearing in the data of $H^m$ and of its differential is $p=(m+1)-a=m-a+1$.

The truncation computes the same cohomology. By [F1], the comparison of $X_{\bullet }$ with its $N$-coskeleton truncation ${\mathrm{cosk}}_N{\mathrm{sk}}_N{X}_{\bullet }$ is the identity through simplicial degree $N$, and the total complex in a fixed total degree uses only finitely many levels. Choosing $N\ge m-a+1$ makes the two total complexes agree in degrees $m-1$, $m$ and $m+1$; hence they have the same $H^m$ and the same augmentation maps there. Since $m$ was arbitrary, every truncation with $N$ at least the stated bound computes the same cohomology $H^m(X,F)$.

Constant coefficients and sharpness for hard truncation. For $F=k_X$ the lower bound is $a=0$, so the actual total-degree computation uses the outgoing map from level $n$ to level $n+1$. To see that omitting that map can change the answer in every degree, take the proper surjection from two reduced points $S=\{0,1\}$ to a point and its Cech nerve. The map is proper since preimages of compact sets are finite, and its fiber powers are finite discrete spaces. A sheaf on a finite discrete space is a family of vector spaces, so global sections are a finite product and exact. Hence the derived section model has internal degree zero only. The normalized degree-$p$ cochains are $k^2$, indexed by the two alternating nondegenerate strings of length $p+1$. Interior deletions give degenerate strings and vanish in the normalized differential; the two endpoint deletions give ${\delta }_p(v_0,v_1)=(v_1+(-1{)}^{p+1}{v}_0,v_0+(-1{)}^{p+1}{v}_1)$, which has rank one over every field. The full augmented complex is contracted by inserting one point, as in [F1], so its positive cohomology is zero and $H^0=k$. If one retains only cochain levels $0,\dots ,n$ and discards ${\delta }_n$, its top cohomology is $k$ for $n>0$ and $k^2$ for $n=0$, giving the wrong answer. Thus level $n+1$ is necessary in general for this hard truncation of the cochain data. A coskeleton truncation is different: it reconstructs higher levels from matching limits, so no claim that ${\mathrm{cosk}}_n{\mathrm{sk}}_n{X}_{\bullet }$ itself fails descent is made.