If \(n\equiv9\mod14\), then there is some integer \(k\) for which \(n=k+\dfrac9{14}\) where \(k\) would be a multiple of 14, i.e. \(k\equiv0\mod14\).
Because \(k\) is a multiple of 14, it's also a multiple of 7, so \(k\equiv0\mod7\).
This then means that \(n\equiv9\equiv2\mod7\), so the remainder upon dividing by 7 is 2.
