$a\equiv\bar a\!\pmod{\!kn}\Rightarrow a\equiv\bar a\!\pmod{\! n};\ $ $(a\bmod kn)\bmod n=a\bmod n.\ $ Congruences persist mod factors of the modulus

The phenomenon you observed holds in greater generality.

Suppose that $m$ and $n$ are positive integers such that $m$ divides $n$. Then for any integer $a$ we have $(a\bmod n)\bmod m= a\bmod m$.

Certainly $(a\bmod n)\bmod m$ is of the right size, between $0$ and $m-1$.

Since $a$ and $(a\bmod n)$ differ by a multiple of $m$, it follows that the remainder when $(a \bmod n)$ is divided by $m$ is the same as the remainder when $a$ is divided by $m$, which is what we needed to show.

Suppose $a\equiv b \mod pq$ and $b \equiv c \mod p$, then we have $$a=rpq+b=rpq+(sp+c)=(rq+s)p+c$$ so that $a\equiv c \mod p$