Understanding a concentration term
Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.
A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at $6.02\times 10^{23}$ atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!
One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of $\ce{^{12}C}$, most common isotope of carbon. This number happens to be Avogadro's Number - $6.02\times 10^{23}$, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.
Now, if you have one mole of a compound, like $\ce{KCl}$, then you know that there is one mole of the $\ce{KCl}$ formula unit in that sample. Since each $\ce{KCl}$ formula unit has one $\ce{K+}$ and one $\ce{Cl-}$, then each mole of $\ce{KCl}$ has one mole of $\ce{K+}$ ions and one mole of $\ce{Cl-}$ ions.
Now, you have your 0.615 M solution, and you only have 230 mL of it. How much $\ce{KCl}$, $\ce{K+}$, and $\ce{Cl-}$ is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:
$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$