What you need is a conversion between the spherical coordinate system and the Cartesian coordinate system. If $r$ denotes the radius of the sphere, $\theta=\text{polar angle}=\frac\pi2-\text{latitude}\in[0,\pi]$ ($0$ = North pole, $\pi$ = South pole) and $\varphi=\text{longitude}\in[0,2\pi)$, then the conversion from spherical coordinates to Cartesian coordinates is given by
\begin{equation}
(r,\theta,\varphi)\mapsto(x,y,z)=(r\sin\theta\cos\varphi,\,r\sin\theta\sin\varphi,\,r\cos\theta).\tag{1}
\end{equation}
To convert back from spherical coordinates to Cartesian coordinates to spherical coordinates, do the following:
\begin{align}
r&=\sqrt{x^2+y^2+z^2},\\
\theta&=\operatorname{acos}(z/r),\\
\varphi&=\operatorname{atan2}(y,x),
\end{align}
where atan2 is the quadrant-aware variation of the arc-tangent function. (Certainly you don't need the first equation in your case, as $r$ is a constant.) If you need to report the longtitude in your software, you should check the range of the implementations of acos and atan2 on your computer. If, e.g., the range of atan2 on your computer is $[-\pi,\pi)$, then you should compute $\varphi$ as the remainder of $\operatorname{atan2}(y,x)+2\pi$ modulo $2\pi$.
If you swing the point upward along a meridian by $15^\circ=\pi/12$ radians, the adusted spherical coordinates become
$$
\begin{cases}
(r,\theta-\frac\pi{12},\varphi)&\text{ if }\, \frac\pi{12}\le\theta\le\pi,\\
(r,\frac\pi{12}-\theta,-\varphi)&\text{ if }\, 0\le\theta<\frac\pi{12}.
\end{cases}
$$
Put these new spherical coordinates into $(1)$, you get the Cartesian coordinates of the adjusted point. Similarly, if you swing the point downward by $15^\circ$ along a meridian, the adusted spherical coordinates become
$$
\begin{cases}
(r,\theta+\frac\pi{12},\varphi)&\text{ if } 0\le\theta\le\pi-\frac\pi{12},\\
(r,2\pi-\theta-\frac\pi{12},-\varphi)&\text{ if } \pi-\frac\pi{12}<\theta\le\pi.
\end{cases}
$$
Taking your example of $R=100$ and $d=1$, you get a central angle of
$$\alpha=2\arcsin\frac{d}{2R}\approx0.573°\approx\frac{2\pi}{628}$$
So your points wouls most closely resemble a regular $628$-gon. But not exactly: the $628$-gon has $d\approx1.0005$ while the $629$-gon has $d\approx0.9989$. Therefore you can't place points at unit distance from one another along a circle of radius $100$ all the way around. Not exactly.
But perhaps you only want points on a part of the circle, so they don't have to get back exactly to their starting point. Or perhaps a slight deviation in radius or distance is acceptable. Then you can get coordinates for your points using
$$ x_k = R\cos(k\alpha) \qquad y_k = R\sin(k\alpha) $$
for $k\in\{0,1,2,\dots,n-1\}$. If you aim for an exact $n$-gon, you'd use $\alpha=\frac{2\pi}n$ while for exact lengths $R$ and $d$ but a non-closing sequence you'd use the angle as computed above.
Best Answer
$(x_k, y_k) =(x_0+r\cos(2k\pi/n), y_0+r\sin(2k\pi/n)) $ for $k = 0 $ to $n-1$.