What is the difference between "margin of error" and "standard error"? Is "margin of error" the same as "standard error"?
A (simple) example to illustrate the difference would be great!
 A: Short answer: they differ by a quantile of the reference (usually, the standard normal) distribution.
Long answer: you are estimating a certain population parameter (say, proportion of people with red hair; it may be something far more complicated, from say a logistic regression parameter to the 75th percentile of the gain in achievement scores to whatever). You collect your data, you run your estimation procedure, and the very first thing you look at is the point estimate, the quantity that approximates what you want to learn about your population (the sample proportion of redheads is 7%). Since this is a sample statistic, it is a random variable. As a random variable, it has a (sampling) distribution that can be characterized by mean, variance, distribution function, etc. While the point estimate is your best guess regarding the population parameter, the standard error is your best guess regarding the standard deviation of your estimator (or, in some cases, the square root of the mean squared error, MSE = bias$^2$ + variance).
For a sample of size $n=1000$, the standard error of your proportion estimate is $\sqrt{0.07\cdot0.93/1000}$ $=0.0081$. The margin of error is the half-width of the associated confidence interval, so for the 95% confidence level, you would have $z_{0.975}=1.96$ resulting in a margin of error $0.0081\cdot1.96=0.0158$.
A: This is an expanded (or exegetical expansion of @StasK answer) attempt at the question focusing on proportions.
Standard Error:
The standard error (SE) of the sampling distribution a proportion $p$ is defined as:
$\text{SE}_p=\sqrt{\frac{p\,(1-p)}{n}}$. This can be contrasted to the standard deviation (SD) of the sampling distribution of a proportion $\pi$:   $\sigma_p=\sqrt{\frac{\pi\,(1-\pi)}{n}}$.
Confidence Interval:
The confidence interval estimates the population parameter $\pi$ based on the sampling distribution and the central limit theorem (CLT) that allows a normal approximation. Hence, given a SE, and a proportion, $95\%$ the confidence interval will be calculated as:
$$p\,\pm\,Z_{\alpha/2}\,\text{SE}$$
Given that $Z_{\alpha/2}=Z_{0.975}=1.959964\sim1.96$, the CI will be:
$$p\,\pm\,1.96\,\sqrt{\frac{p\,(1-p)}{n}}$$.
This raises a question regarding the utilization of the normal distribution even if we really don't know the population SD - when estimating confidence intervals for means, if the SE is used in lieu of the SD, the $t$ distribution is typically felt to be a better choice due to its fatter tails. However, in the case of a proportion, there is only one parameter, $p$, being estimated, since the formula for the Bernouilli variance is entirely dependent on $p$ as $p\,(1-p)$. This is very nicely explained here.
Margin of Error:
The margin of error is simply the "radius" (or half the width) of a confidence interval for a particular statistic, in this case the sample proportion:
$\text{ME}_{\text{@ 95% CI}}=1.96\,\sqrt{\frac{p\,(1-p)}{n}}$.
Graphically,

A: The margin of error is the amount added and subtracted in a confidence interval. 
The standard error is the standard deviation of the sample statistics if we could take many samples of the same size. 
A: sampling error measures the extent to which a sample statistic differs with the parameter being estimated
on the other hand standard error try to quantify the variation among sample statistics drawn from the same population
A: Using @Antoni Parellada's example of sampling proportion, $\hat{p}$,
The standard error of the sample is defined as:
$$
\hat{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
$$ 
The margin of error utilizes the z-score at a specified level of confidence $\alpha$ (e.g., $\alpha =$ 0.05 corresponds to a 95% CI):
$$
M = z_{\alpha/2} \cdot \hat{SE}
$$ 
The margin of error is the half-width of the confidence interval [of the sampling proportion in this case]: 
$$
\hat{p} \pm M 
$$
