All I know that as a conductor in electrostatic equilibrium the field inside it is zero. So if we take a concentric gaussian surface with a radius $b<r<c$ we see that the total charge inside it has to be zero. And since the charge lies on the surface of the shell the inner walls have to be charged with $−Q$.
That is all correct.
Apply Gaussian surfaces in regions 2, 3 and 4. The charged enclosed by surface 2 is $+Q$ so there is a field $E(r)=kQ/r^2$ in the region $a<r<b$. A charge $-Q$ is induced on the inner surface of the conducting shell, so the total charge enclosed by surface 3 is $0$. Hence the electric field in region 3 $(b<r<c)$ - ie inside the conducting shell - is $E(r)=0$.
The total charge on the conducting shell is $-2Q$. There is already a charge $-Q$ on the inner surface so the charge on the outer surface is $-Q$. Surface 4 encloses net charge $-Q$ hence the field in region 4 $(r>c)$ is $E(r)=-kQ/r^2$.
If there were no charged sphere 1 inside the spherical shell, the charge enclosed by both surfaces 2 and 3 would be $0$, so there would be no field in regions 2 or 3 - ie $E(r)=0$ for $0<r<c$.
In both cases the field inside the conductor itself is $0$, regardless of whatever charges are inside or outside the conductor, or on the inner or outer surfaces. The field inside a cavity enclosed by the conductor is only zero if there is no charge in the cavity.
The charge on the conducting shell creates zero electric field in the region $r<b$, so the conducting shell has no effect on the field in region 2 due to the sphere.
Your textbook is applying the Superposition Principle : the field in the various regions is the superposition of the separate fields from the charges when considered separately.
The field in region 2 due to the charge on the shell is zero, but the field in region 2 due to the charge on the sphere is $E(r)=+kQ/r^2$. The total field in this region is the sum of these two fields. The same argument applies for region 4, where the net electric field is $E(r)=+kQ/r^2-2kQ/r^2=-kQ/r^2$.
If we assume that the entire charge of $-2Q$ on the shell remains on its outer surface - which is where it would be if the inner charge did not exist - then by the Superposition Principle the field inside the shell $(b<r<c)$ should be $E(r)=+kQ/r^2$ - which contradicts the principle that the electric field inside a conductor is zero.
The difficulty of applying the Superposition Principle to region 3 (inside the shell) is that the principle assumes that charges do not move when other charges are brought into proximity to them. This is true for the inner sphere - the charge remains on its outer surface when the charged conducting shell is brought to $r=c$ from $r=\infty$. But it is not true for the shell.
If the shell had infinitesimal thickness, there would be no difference between the inner and outer surfaces, so in effect this charge cannot move. But if the shell has finite thickness, charge can move from one surface to the other.
Whether the charge on the shell is on the outer or inner surface makes no difference to the electric field it generates in the cavity inside the shell $(r<b)$, but it does have an effect on the electric field inside the shell itself $(b<r<c)$. The electric field in this region is now the sum of that from the sphere $E(r)=+kQ/r^2$ and that from the charge on the inner surface of the shell $E(r)=-kQ/r^2$. That sum is zero.