If we look at the $x$- and $y$-components of the forces, we see that
$$
\begin{align}
W_x - N_x &= m\omega^2 R\cos\theta,\\
W_y - N_y &= 0.
\end{align}
$$
The normal force $\vec{N}$ is a radial force, so if we assume that the Earth is spherically symmetric, we indeed have $N_x = N\cos\theta$ and $N_y = N\sin\theta$, so that $N=N_R$. But the weight force $\vec{W}$ is not purely radial. It also
has a tangential component, precisely because of the rotation of the Earth:
$$
\begin{align}
W_R &= W_x\cos\theta + W_y\sin\theta,\\
W_\theta &= -W_x\sin\theta + W_y\cos\theta.
\end{align}
$$
So in general $W_x\neq W\cos\theta$. For the normal force, we have
$$
\begin{align}
N_R &= N_x\cos\theta + N_y\sin\theta = N\cos^2\theta + N\sin^2\theta = N,\\
N_\theta &= -N_x\sin\theta + N_y\cos\theta = 0.
\end{align}
$$
From
$$
\begin{align}
W_x\cos\theta - N_x\cos\theta &= m\omega^2 R\cos^2\theta,\\
W_y\sin\theta - N_y\sin\theta &= 0,
\end{align}
$$
we find the centripetal force
$$
W_R - N = m\omega^2 R\cos^2\theta,
$$
but note that there's also a tangential force
$$
W_\theta = -m\omega^2 R\sin\theta\cos\theta.
$$
Okay. Firstly, I would like to point out that you are mixing two very different concepts here:
(1) Variation in the value of gravity $g$ as the distance from the surface of the earth changes.
(2) True and apparent weight
(1) Variation in the value of gravity
Alright. Variation in gravity. Firstly, lets get clear on the value of $g$. What exactly is $g$? It's like this: Suppose you are somewhere. Maybe sitting somewhere having pizza or flying in the sky. The earth applies a force on you. Let's call this force $F$. Then the value of $g$ is simply defined as $F/m$. That's it.
Now suppose the radius of earth is $R$ and you are at distance $d$ from the surface. (Note, from surface of the earth, not the center.) The force applied on you by the earth is
$$F = \cfrac{GM_em}{(R+d)^2}$$
So, now,
$$g = F/m = \cfrac{GM_e}{(R+d)^2}$$
Have a look at it. The value of $g$ indeed depends on $d$, your distance from the surface of the earth. But, near the surface of the earth, $d<<R$, so we can approximate the above expression to
$$g = F/m = \cfrac{GM_e}{R^2}$$
which is independent of $d$. But note that it is valid only for small values of $d$.
(2) True and apparent weight
Okay. Answer to the next part of the question. True and apparent weight. True weight is simply weight. What is your true weight? It's simply $mg$. Mass multiplied by gravity. End of story.
Now, Apparent weight. I'll denote it by $W_A$. It's defined as
$$W_A = N$$
where $N$ is the normal force in the direction opposite to the direction of gravity. That is away from the center of the earth. You may be standing and someone may be trying to push you horizontally. That normal reaction force doesn't count. Only the vertical Normal Force counts.
So suppose you jump from the top of the building because your dog died. You are falling. Your '(True) Weight' is simply $mg$. Your Apparent weight is $0$. Because there is no normal force applied on you currently. (Offcourse the ground will apply one hell of a normal force when you finally reach it.)
Now suppose you are standing in an elevator at rest. True weight, offcourse is $mg$. But Apparent weight is also $mg$. Because you are at rest, $N = mg$.
Elevator moving with constant speed: $N = mg$
Suppose the magnitude of elevator's acceleration is $|a|$.
Elevator moving upwards, and slowing down: $N = mg - m|a|$
Elevator moving upwards, and increasing speed: $N = mg + m|a|$
Elevator moving downwards, and slowing down: $N = mg + m|a|$
Elevator moving downwards, and increasing speed: $N = mg - m|a|$
So, why did they introduce the concept of Apparent Weight. Apparent weight is the weight you 'feel'. Think about it! When you are falling, you feel weightlessness. Hence Apparent Weight is $0$. When in an elevator with moving upwards with increasing speed, you feel heavier. Hence more is the Apparent Weight!
(3) $d/R$ ratio
The ratio $d/R$ where weight would be 1% lesser:
$$\cfrac{GM_e}{(R+d_1)^2} = 0.99 \cfrac{GM_e}{R^2}$$
Solve it for $d_1$. That's your answer!
Best Answer
A weighing machine measures the force exerted by a body on the weighing machine.
Newton's third law then predicts that there is a force of the same magnitude and opposite in direction acting on the body producing the force.
On the Earth if the weighing machine and the body are not accelerating (ignoring the rotation of the Earth) then the reading on the weighing machine will be the weight of the body.
If the weighing machine and the body are accelerating then you could call the reading on the weighing machine the apparent weight of the body.
So including the effect of the rotation of the Earth it is only at the geographic poles that reading on the weighing machine is the weight of the body.
Elsewhere on the Earth the reading on the weighing machine will be lower than at the poles so you could call that the apparent weight.
The difference between these readings is small.
If the weight of the body is $10 \, \rm N$ then with the weighing machine and the body in a stationary lift, or a lift moving at constant velocity upwards or downwards the reading on the weighing machine would be $10 \, \rm N$ which is the weight of the body.
If the weighting machine and the lift had an upward acceleration of $5 \,\rm m s^{-2}$ then the reading on the weighing machine would be $15 \, \rm N$ and you could say that the apparent weight of the body was $15 \, \rm N$
If the weighting machine and the lift had a downward acceleration of $5 \,\rm m s^{-2}$ then the reading on the weighing machine would be $5 \, \rm N$ and you could say that the apparent weight of the body was $5 \, \rm N$
If the weighting machine and the lift had a downward acceleration of $10 \,\rm m s^{-2}$ then the reading on the weighing machine would be $0 \, \rm N$ and you could say that the apparent weight of the body was zero - the body appeared to be weightless.
The definition of weight that I have used is that the weight of a body is the force on the body due to the gravitational attraction of the Earth.
However others define the weight of a body as the reading on a weighing machine as explained by Walter Lewin in one of his 8.01 Classical Mechanics lectures.
Using this definition a body is weightless when it is in free fall.