$$\text{s.d.}(\bar{x}) = \frac{\sigma}{\sqrt{n}}$$
This is approximately the average distance of the possible \(\bar{x}\) values (for repeated samples of size \(n\)) from the true population proportion \(\mu\).
The issue here is that you typically do not know \(\sigma\) in advance.
Instead, use the sample standard deviation \(s\) to estimate the population standard deviation.
$$\text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}}$$
This estimates the difference approximately.
$$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$$
However, the dilemma is that you rarely know sigma. So you replace the standard deviation with the standard error.
$$z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$$
This notably does not have a normal distribution. Instead, it has a t-distribution with degrees of freedom of \(n-1\).
Every statistic has a sampling distribution, but the distribution may not always be normal or bell-shaped.
An interval is a range of reasonable values for the parameter with a high level of confidence.
The Central Limit Theorem states that, as the sample size gets large, the sampling distribution approaches:
$$N(\mu, \frac{\sigma}{\sqrt{n}})$$
This makes the confidence interval:
$$\bar{x} \pm t^* \text{s.e.}(\bar{x})$$
The rule of thumb is that, if you're finding a CI for a proportion, use a \(z\). For a mean, use a \(t\). If the sample size is large (n > 30), then it begins to approximate the normal distribution. Make sure to always round down the degrees of freedom, if there isn't a match. That gives you a large bound of error.