$2$-dimensional rotations are enough to understand all of this.
You said you know that $(\cos\phi,\sin\phi)^\top$ describes a circle in $2$D and you want to consider many rotated copies of this circle so that it formes a sphere. The first thing we have to do is to go to higher dimensions. So when our circle is placed in the $xy$-plane of out $3$-dimensional space, it is given by
where we just added a zero in the last component to show that it has no expanse into the $z$-direction.
$3$-dimensional rotation matrices
You said you know $2$-dimensional rotation matrices. Here is how to build a simple $3$-dimensional equivalent:
While in two dimensions there is no axis to rotate on (we rotate around points), in three dimensions we have to specify such an axis. This example is very simple. Our modified matrix just has another row/columns which shows that we completely ignore the point's $z$-coordinate. We rotate as if it where a $2$D point in the $xy$-plane. After rotation the point has the same $z$-coordinate as before the rotation (because of the $1$ in the $zz$-entry of the matrix).
So the matrix $R_z(\phi)$ rotates around the $z$-axis by an amount $\phi$. Here are the matrices which rotate around $y$- and $x$-axis respecitively (I think you can see why):
The rotated circle
Now we have our $3$D circle in the $xy$-plane around the $z$-axis. To make it rotate in such a way that it forms a sphere, we must rotate it around an axis other than $z$, e.g. the $x$-axis. So we take out circle-description $(\cos\phi,\sin\phi,0)^\top$ and we apply a rotation matrix $R_x(\theta)$:
Well, this last vector might not be exactly what was given to you by the other answer, but this may be the case because they chose a circle in an other plane than the $xy$-plane or rotated around an other axis than $z$. You can play a bit with this idea to find the correct configuration. At least, this result shows structural similarities to the one given to you in the other answer.