Suppose $P = (X, Y, Z)$ is a randomly chosen point on the surface of such a sphere. Would you agree that the marginal cumulative distribution function of $X$ would be $$\Pr[X \le x] = \frac{\text{surface area of "cap" with } X \text{-values at most } x}{\text{total surface area}}$$ for some $x \in [-R, R]$? If so, then recall that the surface area of such a cap is given by $$A(x) = 2\pi R (x+R).$$ Hence $$F_X(x) = \Pr[X \le x] = \frac{2\pi R (x+R)}{4 \pi R} = \frac{x+R}{2R}.$$ Therefore $$f_X(x) = \frac{\mathbb 1(-R \le X \le R)}{2R};$$ that is to say, $X$ has uniform density on $[-R,R]$ as claimed.
Why does the calculation you proposed not work? The reason is because you are comparing an arclength (the circumference measure) against an area. The measures of these are not the same. In particular, by making such an argument, you are ignoring a factor that relates the particular $x$-value to the probability of being in a neighborhood of that $x$.
What I mean by this is that you need to be more careful about formalizing the argument. Consider instead the limiting behavior as $\epsilon \to 0$ of the probability $$\Pr[x - \epsilon \le X \le x + \epsilon].$$ Then instead of a circumference, we are interested in a thin "band" of the sphere's surface satisfying $|X - x| \le \epsilon$. The first thing you will notice is that for a fixed choice of $\epsilon > 0$, the lateral width of the band is not constant as a function of $x$: for instance, when $x$ is "close" to $R$, the band will appear thickest, compared to $x$ "close" to $0$, where the band appears thinnest, even though the band is sandwiched between planes equally spaced apart by $2\epsilon$. Therefore, you must adjust the calculation accordingly, in as much as computing the arclength of a circular sector does not give the same result as the projection of that arclength onto the $x$-axis.
