4 added 449 characters in body

EDIT: reading your question again, I'm a bit confused about both terminology and what you are actually asking. Below by an exact complex I mean one with vanishing cohomology (as a complex!), whereas by an acyclic complex I mean a complex of ($\Gamma$-)acyclic objects. In general, the hypercohomology of an acyclic complex is easy to compute (see first part of my answer), and the hypercohomology of an exact complex is zero (see second half).

To elaborate on Damian's comment (as Sándor Kovács points out, the second half is wishful thinking):

One way to define hypercohomology of a complex $K^\bullet$ is to take a Cartan-Eilenberg resolution $K^{\bullet, \bullet}$ and then take the cohomology of the total complex of $K^{\bullet, \bullet}$. Cartan-Eilenberg resolutions are pretty awesome (they have basically all nice properties you could ask for), but in particular every columns is an injective resolution of the respective term of $K^\bullet$.

Now given any left-bounded double complex $K^{\bullet, \bullet}$ there exist two spectral sequences converging to the cohomology of the total complex (see Weibel). The $E_2$ pages are obtained by taking first horizontal and then vertical cohomology, and vice versa.

In what follows, I shall write $h^p(K^\bullet)$ for the cohomology of the complex $K^\bullet$, and $H^p(K)$ for the derived functor of global sections on $K$.

In our case, taking vertical cohomology first, we get an $E_1$ page $H^q(K^p)$. By your acyclicity assumption, we have $H^q(K^p) = 0$ for $q>0$, i.e. the $E^1$ page just consists of $K^\bullet$ again in the lowest row, and zeros else. Thus the spectral sequence degenerates at the $E_2$ page, leaving you with $h^p(\Gamma(K^\bullet))$ for the cohomology of the total complex, i.e. hypercohomology (see Weibel again for how the separate pages relate to the cohomology of the total complex).

This is the statement you were asking for (to the extent that I am aware of a correct formulation): the hypercohomology of an acyclic complex is just the cohomology of its complex of global sections. In particular an exact acyclic complex has vanishing hypercohomology.

More generally, any exact complex has vanishing hypercohomology: this time look at the spectral sequence where we take horizontal cohomology first. Another awesome property of cartan-eilenberg resolutions: the horizontal boundaries and cycles are themselves injectives, and hence taking horizontal cohomology of the cartan-eilenberg resolution yields an injective resolution of the cohomology of $K^\bullet$. There is a small difficulty: we are not supposed to compute the hypercohomology of $K^{\bullet,\bullet}$ but of $\Gamma(K^{\bullet,\bullet})$. However, since injectives are acyclic, "taking horizontal cohomology" and "taking global sections" actually commutes in our case, so we are good. In summary: there exists a convergent spectral sequence with $E_2$ page $H^q(h^p K^\bullet)$, which is evidently zero if $K^\bullet$ is exact.

Even more generally (and by a very similar argument), hypercohomology factors through quasi-isomorphism.

Note: again as Damian points out, "hypercohomology on some abelian category" does not really make sense. You can in general define hyperderived functors, and then hypercohomology is usually the name for the hyperderived functor of global sections. My above statements hold for the hyperderived functors of an arbitrary left-exact additive functor $\Gamma$.

3 added 1196 characters in body

In what follows, I shall write $h^p(K^\bullet)$ for the cohomology of the complex $K^\bullet$, and $H^p(K)$ for the derived functor of global sections on $K$.

In our case, taking vertical cohomology first, we get an $E_1$ page $H^q(K^p)$. By your acyclicity assumption, we have $H^q(K^p) = 0$ for $q>0$, i.e. the $E^1$ page just consists of $K^\bullet$ again in the lowest row, and zeros else. Thus the spectral sequence degenerates at the $E_2$ page, leaving you with $h^p(\Gamma(K^\bullet))$ for the cohomology of the total complex, i.e. hypercohomology (see Weibel again for how the separate pages relate to the cohomology of the total complex).

This is the statement you were asking for (to the extent that I am aware of a correct formulation): the hypercohomology of an acyclic complex is just the cohomology of its complex of global sections. In particular an exact acyclic complex has vanishing hypercohomology.

More generally, any exact complex has vanishing hypercohomology: this time look at the spectral sequence where we take horizontal cohomology first. Another awesome property of cartan-eilenberg resolutions: the horizontal boundaries and cycles are themselves injectives, and hence taking horizontal cohomology of the cartan-eilenberg resolution yields an injective resolution of the cohomology of $K^\bullet$. There is a small difficulty: we are not supposed to compute the hypercohomology of $K^{\bullet,\bullet}$ but of $\Gamma(K^{\bullet,\bullet})$. However, since injectives are acyclic, "taking horizontal cohomology" and "taking global sections" actually commutes in our case, so we are good. In summary: there exists a convergent spectral sequence with $E_2$ page $H^q(h^p K^\bullet)$, which is evidently zero if $K^\bullet$ is exact.

Even more generally (and by a very similar argument), hypercohomology factors through quasi-isomorphism.

2 edited body; edited body; added 157 characters in body

To elaborate on Damian's comment (as Sándor Kovács points out, the second half is wishful thinking):

One way to define hypercohomology of a complex $K^\bullet$ is to take a Cartan-Eilenberg resolution $K^{\bullet, \bullet}$ and then take the cohomology of the total complex of $K^{\bullet, \bullet}$. Cartan-Eilenberg resolutions are pretty awesome (they have basically all nice properties you could ask for), but in particular every columns is an injective resolution of the respective term of $K^\bullet$.

Now given any left-bounded double complex $K^{\bullet, \bullet}$ there exist two spectral sequences converging to the cohomology of the total complex (see Weibel). The $E_2$ pages are obtained by taking first horizontal and then vertical cohomology, and vice versa.

In our case, taking vertical cohomology first, we get an $E_1$ page $H^q(K^p)$. By your acyclicity assumption, we have $H^p(K^q) H^q(K^p) = 0$ for $p>0$, q>0$, i.e. the$E^1$page just consists of$K^\bullet$again in the lowest row, and zeros else. Thus the spectral sequence degenerates at the$E_2$page, leaving you with$h^p(\Gamma(K^\bullet))$for the cohomology of the total complex, i.e. hypercohomology (see Weibel again for how the separate pages relate to the cohomology of the total complex). This is the statement you were asking for (to the extent that I am aware of a correct formulation): the hypercohomology of an acyclic complex is just the cohomology of its complex of global sections. Note: again as Sándor Damian points out, "hypercohomology on some abelian category" does not really make sense. You can in general define hyperderived functors, and then hypercohomology is usually the name for the hyperderived functor of global sections. My above statements hold for the hyperderived functors of an arbitrary left-exact additive functor$\Gamma\$.

1