The statement $a \mod{b} = c \mod{b}$ is equivalent to saying that there exists some integer $r$ such that $a-c=rb$. So in particular if $a - (a+kb)=kb = k \cdot b$ hence $a \mod{b} = a+kb \mod{b}$ is always true for any integer $k$.
For your second part, we want to know what is the least $K$ such that $xK \mod{y} = 0 \mod{y}$ for some given integers $x,y$. Or to rephrase that, we want to know what is the least $K$ such that there exists an integer $r$ with $xK=ry$. Clearly $K=y$ works as we can then take $r=x$ but there's no guarantee that this is the least possible solution.
Indeed, consider the case $x=4$ and $y=6$. Then $K=y=6$ works, but we could have taken $K=3$ (with $r=2$) as $4 \times 3 = 6 \times 2$.
Looking at this more closely, it turns out then whenever $x$ and $y$ share a common divisor, we can take this factor out of $y$, so in the end the least possible value is $K=\frac{y}{\gcd(x,y)}$ (this is always an integer).
To see that this works, note that $xK=\frac{xy}{\gcd(x,y)}=y\frac{x}{\gcd(x,y)}$ so $r=\frac{x}{\gcd(x,y)}$. I'll leave you to check that this is actually minimal.
