Hello Long, and welcome to PF :)
Slightly irregular: a problem on the homework forum that the poster isn't solving. No wonder, because the solution is provided in the text.
In appreciation of your sense of wonder, a few questions/remarks to get you thinking a bit more:
"Why doesn't left loop happen and right loop be (do?) nothing ?" reduces to "Why doesn't right loop do nothing ?", because left loop really "happens".
The closing of S means there is zero resistance between its connection points, so zero voltage. At t=0 the capacitor is charged to a voltage ##\epsilon## and between its connection points it "sees" S and R2 in series. That means it starts to discharge ! Initially with a current ##\epsilon##/R2, but while it discharges, the voltage over the capacitor drops. (well, continue reading after "according to..".
So through switch S flow both the currents from the left and from the right circuit. Being a good conductor (zero resistance) it has no problem with that.What would you say to a classmate who is asking "Why doesn't right circuit happen and left circuit do nothing ?"
What about a second switch, S2, parallel to S, that is closed at the same time as S ? Would that make a difference in your perception ? And in reality ?