The concept that relates $F = ma$ and its rotational counterpart $Fr = Ia$ is rotational symmetry.
More generally, rotational symmetry is the general relation between linear mechanics and rotational mechanics.
$F = ma \quad $ for linear acceleration
$Fr = I \alpha \quad $ for angular acceleration
Obviously, the minimum number of spatial dimensions for rotation to be possible at all is 2.
Our space has three spatial dimensions, so a general treatment should cover three spatial dimensions, but a discussion of the simplified case of two spatial dimensions is sufficient to establish the relation between linear mechanics and rotational mechanics.
So: I simplify the case to two spatial dimensions.
For the purpose of this discussion the simplest object is a (hula) hoop. A hoop with radius $r$
We can think of that hoop as divided in subsections, we can make those subsections as small as we want. We can think of a force that causes angular acceleration of that hoop as distributed along the entire perimeter of that hoop. That is, we can think of each subsection of the hoop as being subjected to a force that acts on that particular subsection. We set up this demonstration in such a way that that at every point along the perimeter the force is always tangential to the local perimeter. We can think of an accumulated version of the force as a summation over all the subsections.
The point of this setup is to capitalize on rotational symmetry.
For the purpose of calculation you can treat the hoop as a hoop with radius $r$.
Or: you can with the same result treat this case as if all of the mass is in a single point, moving along at a fixed distance $r$ to a center of rotation (in this case the center of mass of the hoop), subject to the accumulated force.
I will of course express the angular acceleration in radians per second squared.
The relation between linear acceleration and angular acceleration is of course the same as the relation between linear velocity $v$ and angular velocity $\omega$.
$ a = \alpha r \qquad (1)$
The local acceleration tangential to the circle can be expressed as a linear acceleration.
$F = ma \qquad (2) $
Replacing linear acceleration $a$ with angular acceleration $\alpha$
$F = m\alpha r \qquad (3) $
Adding a factor $r$ for 'radius' on both sides so that the left hand side of the equation expresses torque:
$Fr = m\alpha r r \qquad (4) $
$Fr = mr^2\alpha \qquad (5) $
(5) contains a factor $mr^2$. At this point it is a natural step to define a concept of moment of inertia: $I$, defining its magnitude as $mr^2$.
$Fr = I\alpha \qquad (6) $
The necessary condition for this reasoning to hold good is that the force you are dealing with behaves the same in all spatial orientations. In other words, the necessary condition is rotational symmetry.