A planet has mass $M$, radius $r$ and orbital period $\tau$.
Then the force per unit mass exerted on an object at rest at the equator on the surface of the planet is given by $$\frac{GM}{r^2} - \frac{4\pi^2 r}{\tau}$$
I am unsure why these should be subtracted?
Does the gravitational force not act in the same direction as the centripetal force? Also, what is the significance of the object being at rest of the equator? Would the result be different?
Yes, the gravitational force acts in the same direction as the centripetal force because gravity is the centripetal force that's pulling things down to the center of the planet. Without gravity, everything on the surface of a rotating planet would continue moving in a straight line and thus fly off into space.
In particular, in an inertial reference frame the only forces acting on your unit mass object (assuming a spherical airless planet) are the gravitational force of magnitude $GM/r^2$, which pulls the object towards the center of the planet, and the ground reaction force exerted by the planet's surface that keeps the object from sinking into it.
These two forces are obviously opposite in direction, and we can calculate what the magnitude of the ground reaction force needs to be in order to keep the object moving in a circular trajectory around the center of the planet at radius $r$ and period $\tau$. In particular, to maintain this circular trajectory, the net centripetal force (i.e. the difference between the gravitational and ground reaction forces) per unit mass acting on the object must equal $$\frac{F_g - F_s}m = a = \omega^2r = \left(\frac{2\pi}{\tau}\right)^2r = \frac{4\pi^2r}{\tau^2},$$ where $F_g$ and $F_s$ denote the magnitudes of the gravitational and the ground reaction forces acting on the object and $a$ denotes the centripetal acceleration needed to maintain the object on a circular trajectory at radius $r$ with angular velocity $\omega = 2\pi/\tau$.
Thus, given the known magnitude $F_g/m = GM/r^2$ of the gravitational force per unit mass, we can calculate that the ground reaction force preventing the object from sinking into the planet must have a magnitude of $$\frac{F_s}m = \frac{F_g}m - \frac{4\pi^2r}{\tau^2} = \frac{GM}{r^2} - \frac{4\pi^2r}{\tau^2}.$$
However, it seems (from the fact that the object is described as being "at rest") that your exercise is not assuming an inertial reference frame, but rather a rotating reference frame matching the rotation of the planet. In such a reference frame the object lying on the planet's surface is not following a circular trajectory, but rather appears to be motionless with respect to the frame of reference (and the planet).
Being motionless (and thus not accelerating), the object should have no net force acting on it. Yet we know that the gravitational and ground reaction forces haven't actually changed.
The way to resolve this apparent contradiction (and other similar contradictions regarding non-inertial reference frames) is to introduce a centrifugal pseudo-force that exactly makes up the difference. In particular, for an object at radius $r$ from the center of rotation in a rotating reference frame, the magnitude of the centrifugal (pseudo-)force per unit mass is $$\frac{F_{cf}}m = \omega^2r = \frac{4\pi^2r}{\tau^2},$$ i.e. exactly the same as the force per unit mass required to keep the object moving on a circular trajectory at angular velocity $\omega = 2\pi/\tau$ at radius $r$ around the center of rotation.
As its name indicates, the centrifugal (pseudo-)force points away from the center of rotation, i.e. away from the planet's center in this case. Thus, the net force per unit mass acting on the object in this rotating reference frame, including pseudo-forces arising from the rotation of the reference frame, is $$\frac{F_g - F_s - F_{cf}}m = 0,$$ just as it needs to be in order for the object to remain motionless.
(If the object were not stationary with respect to the reference frame, we'd also have to include a Coriolis pseudo-force as an additional correction to make the forces sum to zero. But for a stationary object the Coriolis force is always zero.)
Conversely, we can also use this equation to calculate the magnitude of the ground reaction force $F_s$ in the rotating coordinate system, since the gravitational force $F_g$ and the centrifugal pseudo-force $F_{cf}$ are known based on the given parameters of the exercise. It works out exactly to the same value of $$\frac{F_s}m = \frac{GM}{r^2} - \frac{4\pi^2r}{\tau^2}$$ as we calculated in the inertial reference frame, which is exactly as it should be: being an actual physical force mediated by short-range electromagnetic interactions, the surface reaction force should not be dependent of the choice of reference frame.
Ps. See these answers I wrote earlier for more discussion on centrifugal and other "fictitious" forces in rotating coordinate systems, including diagrams:
I couldn't figure out a good way to work those links into the text of the answer above, but I still feel they may be useful further reading, so I'll just stick them in a postscript here.
The gravitational force is directed towards the center of the planet while the centrifugal force (second term) is directed outward. Hence the minus sign. If the object were not at rest then the centrifugal force would not be as in the formula.
If the planet $2\pi r$ meter in $\tau$ seconds, the velocity for a fixed point on the equator is $\frac{2\pi r}{\tau}$. So the centrifugal force is $\frac{mv^2}{r}=\frac{4\pi r}{\tau}$. This force is directed outward, just like the centrifugal force while standing inside a rotating cylinder tends to pull you outside the cylinder. And its magnitude will be only as calculated for a point moving with $v$ along the equator.
A simple free-body diagram would show that $$F_N-\frac{GMm}{r^2}=ma_c=m\left(-\frac{\omega^2}{r}\right)$$.
The net acceleration is toward the "center" of earth (approx.), and that is negative in my chosen frame if the normal force is positive. Your quantity, force per unit mass, is the contact/normal force divided by the object mass: $$\frac{F_N}{m}=\frac{GM}{r^2}-\left(\frac{2\pi}{\tau}\right)^2 r.$$
Free-body diagrams, properly drawn are powerful things and help prevent "guessing" about adding or subtracting forces.
