As the number of attempts of a binomial distribution approaches infinity, it approximates to a Normal distribution with $\mu= np $ and $\sigma= np(1-p)$.
My question is: When do you use each one? I seem to be getting confused as to whether I should be using
$$\Pr(X\lt x)= \Pr\left(Z\lt \frac{x-\mu}{\sigma}\right) \text{ or } \Pr\left(Z\lt \frac{x-\mu n}{\sqrt{n}\cdot\sigma}\right)$$
On
You only need to use a normal approximation to a binomial distribution if it's too difficult to compute the probability using the binomial distribution. What constitutes "too difficult" depends on what computational resources you have available. With a computer and good mathematical software, calculating binomial probabilities for $n$ in the thousands is no problem at all.
So first of all, DeMoivre-Laplace is the special case of the central limit theorem for the binomial distribution. You don't really need to remember it as a separate result.
Secondly, the point is to do a shift and rescaling so that your random variable has mean zero and variance $1$. For a sum of independent identically distributed random variables, that winds up being $\frac{\sum_{i=1}^n X_i - n\mu}{\sqrt{n} \sigma}$. To rewrite this in terms of the sample mean, you divide top and bottom of that by $n$ to get $\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}$. The binomial distribution is a sum of independent Bernoulli random variables, so if you are estimating a binomial probability you would want the first form.