For largely historical reasons, our time-keeping system is reckoned by the position of the Sun. We can also choose to keep time by reckoning how long it takes for a specific star on the Celestial Sphere to get back to the same point in the sky night after night. Since one rotation of the Earth is all it takes to do this for distant stars, that is the Sidereal Day. NOTE Since one rotation of the Earth also moves the Earth $~1/365^{th}$ of the way around the Earth's orbit around the Sun, the Earth must rotate a little further to get the Sun back into the same position.

That $~1^\circ$ extra (did you catch why $~1^\circ$?) amounts to 4 minutes of extra rotation. It is on this 4 extra minutes of rotation that we base our 24h clock. So the Sidereal Day is 4 minutes shorted than the Solar Day (23h 56m). We still use the 24 hour clock, so we must account for this discrepancy.

Since RA is divided into $24^h$, one might assume that stars travel through our LHCS sphere in $360/24 = 15^\circ/hr$ or (equivalently) $15''/s$. However, the $24^h$ of RA is not quite right. It should be $23^h56^m$ for the Sidereal Day. This comes out to $23*3600 + 56*60 = 86160s$. So our earlier estimate of $15/s' is off by a factor of (24*3600)/86160.

Accounting for this discrepancy, we find that the rate at which objects move through our LHCS is $15*(86400/86160) = 15.0417 /s$.