Why are wheels made to carry more mass on the circumference? We assume an external force $F$ parallel to the horizontal surface on the top edge of a cylindrical wheel with radius $r$ and mass $m $ and moment of inertia $I$.
For this cylinder to roll without slipping it should satisfy the condition:
$a = \alpha* r$ ————(1)
(Where $a$ is the translational acceleration and $ \alpha$ is the angular acceleration.
The friction ($f_s$) acts to balance the changes in a manner so that condition of rolling is met. First, it enhances the net external force ($F + f_s$) and hence the translational acceleration ($a$). Second, it constitutes a torque in anticlockwise direction inducing angular deceleration.
Applying Newton's second law for translation, the linear acceleration of the center of mass is given by :
$ a = \frac{F + f_s}{m}$ ————(2)
Similarly applying Newton's second law for rotation, the angular acceleration of the center of mass is given by:
$ \alpha = \frac{r*(f_s - F)}{I}$ ————(3)
Combing eqn 1,2 and 3 we get the expression for $f_s$ :
$f_s = (\frac{mr^2 - I}{mr^2 + I}) * F $ 
The source says:

For ring and hollow cylinder, $I = mr^2$. Thus, friction is zero even for accelerated rolling in the case of these two rigid body. This is one of the reasons that wheels are made to carry more mass on the circumference.


Now the part I don’t understand is that why do we want to reduce the friction here since eqn 2 says that more friction means more horizontal acceleration which is good for wheels I guess. Help will be really appreciated..
 A: Yes, for your second equation increasing $f_s$ means increasing $a$, but you are blindly applying equations here instead of thinking about the physics.
You have imposed rolling without slipping conditions. Therefore, your equation $f_s = (\frac{mr^2 - I}{mr^2 + I}) * F$ gives you the required static friction force needed to prevent slipping. The larger this value is, the more static friction you need to prevent slipping. So you want this to be smaller. Making this $0$ means that you don't need static friction to prevent slipping, and hence applying your force to the top of the ring is sufficient to cause rolling without slipping. So, by increasing $I$ you are not "decreasing static friction". You are just decreasing what you need the static friction to be so slipping doesn't occur.
As a concrete example, you could push a ring on ice in this manner and there would not be slipping between the ice and the ring; the resulting translational motion from the applied force and resulting rotational motion from the applied torque ends up satisfying the rolling without slipping condition without requiring any additional help from static friction.$^*$
Contrast this to an example where $(\frac{mr^2 - I}{mr^2 + I}) * F>\mu_sN$. Then you can never get rolling without slipping because your required static friction force is larger than the maximum value it can have.

So this means that more the moment of inertia more the “grip” on the road (which is favoured in wheels).

No. Notice that none of your analysis has taken into account the material properties between the two surfaces. What you are essentially doing in your analysis is finding what static friction needs to be in order for rolling without slipping to occur. 
A better way to look at this is to think of rolling without slipping as a "balance" between rotational and translational motion. We need these two types of motion to be related exactly right so that $v_\text{COM}=\omega r$. The moment of inertia is important here because that influences the rotational motion. 
Where the "grip" comes into play is when you are comparing the required static friction force to the maximum value it could obtain for the materials in question.

$^*$Note that the same thing can be done with a solid cylinder if you apply the force half way between the center and the edge of the cylinder. In general if you apply a force a distance $\beta R$ (with $0\leq\beta\leq1$) then the required static friction force to prevent slipping is
$$f_s=\frac{\beta mr^2-I}{mr^2+I}F$$
