At the risk of stating the obvious, perhaps it’s also worth addressing the final question, of “why one needs to take Čech cohomology on the left”, with an example.

The typical object of study in the famous Bestvina—Mess paper that the OP cites is a hyperbolic group together with its Gromov boundary. The Gromov boundary is a fractal (all I mean by this is that it needn’t have the homotopy type of a CW complex) and so the Čech and singular cohomologies needn’t coincide.

In the simplest concrete example, the hyperbolic group $G$ is the free group of rank 2, and $A=\partial G$ is the Cantor space. As mentioned in the question, the fact the OP asks about is used to prove Bestvina—Mess’s Corollary 1.3(b), which states (using $\mathbb{Z}$ coefficients, say):

$H^i(G,\mathbb{Z}G)\cong \smash{\check{H}}^{i-1}(\partial G)$.

If we could get away with singular cohomology, then applying this to the example with $i=1$, we would conclude that the singular cohomology of the Cantor space was a countable direct sum $\bigoplus_i \mathbb{Z}$. But this is (very) false: the 0th singular cohomology of the Cantor space is an uncountable direct product of copies of $\mathbb{Z}$.

This is why it is vital to use Čech cohomology when working with the Bestvina—Mess theorem in full generality.

At the risk of stating the obvious, perhaps it’s also worth addressing the final question, of “why one needs to take Čech cohomology on the left”, with an example.

The typical object of study in the famous Bestvina–Mess paper that the OP cites is a hyperbolic group together with its Gromov boundary. The Gromov boundary is a fractal (all I mean by this is that it needn’t have the homotopy type of a CW complex) and so the Čech and singular cohomologies needn’t coincide.

In the simplest concrete example, the hyperbolic group $G$ is the free group of rank 2, and $A=\partial G$ is the Cantor space. As mentioned in the question, the fact the OP asks about is used to prove Bestvina–Mess’s Corollary 1.3(b), which states (using $\mathbb{Z}$ coefficients, say):

$H^i(G,\mathbb{Z}G)\cong \smash{\check{H}}^{i-1}(\partial G)$.

If we could get away with singular cohomology, then applying this to the example with $i=1$, we would conclude that the singular cohomology of the Cantor space was a countable direct sum $\bigoplus_i \mathbb{Z}$. But this is (very) false: the 0th singular cohomology of the Cantor space is an uncountable direct product of copies of $\mathbb{Z}$.

This is why it is vital to use Čech cohomology when working with the Bestvina–Mess theorem in full generality.