You do it the same way we do with numbers: find a common denominator and then just add the numerators.
For example, if you needed to add
$$\frac{1}{2}+\frac{1}{3}$$
you need to find a "common denominator" (a number which is a multiple of both $2$ and $3$), convert the two fractions to the common denominator, and then add them. The simplest common denominator is just the product of the two denominators, we one way to do this is:
$$\begin{align*}
\frac{1}{2}+\frac{1}{3} &= \frac{1}{2}\cdot\frac{3}{3} + \frac{1}{3}\cdot\frac{2}{2}\\
&= \frac{3}{6} + \frac{2}{6}\\
&= \frac{3+2}{6}\\
&= \frac{5}{6}.\end{align*}$$
So do the same thing here:
$$\begin{align*}
\frac{4x+4h}{x+h+1} - \frac{4x}{x+1} &= \frac{4x+4h}{x+h+1}\cdot\frac{x+1}{x+1} + \frac{4x}{x+1}\cdot\frac{x+h+1}{x+h+1} \\
&= \frac{(4x+4h)(x+1)}{(x+h+1)(x+1)} - \frac{4x(x+h+1)}{(x+h+1)(x+1)}.\end{align*}$$
Now you can just add them, since they have the same denominator, by keeping the denominator and adding numerators. Do the algebra in the numerator, and simplify as needed.
Note: If you are more comfortable with the algebra, there are things you can do to simplify your life; when adding fractions, sometimes the product of the denominators is not the best common denominator: if you can spot a simpler one, use it. For instance, if we had $\frac{1}{6}+\frac{1}{10}$, then we could use $30$ as the common denominator, instead of $6\times 10 = 60$, which leads to smaller numbers, easier arithmetic; likewise, if we had
$$\frac{1}{x^2-1} + \frac{2}{x^2+2x+1}$$
then you could use $(x^2-1)(x^2+2x+1)$ as a common denominator, but if you could spot that $x^2-1 = (x+1)(x-1)$ and $x^2+2x+1 = (x+1)^2$, then you might realize that you can use $(x+1)^2(x-1)$ instead, which is smaller (degree 3 instead of degree 4).
In your situation, on possible simplification is to note that both fractions are multiples of $4$. So you can factor out $4$ and deal with simpler fractions if you want. But if you are unsure about these kinds of simplifying steps, then don't do them. They are not strictly necessary, though they are useful when dealing with more complicated expressions.
There are two confusions here. One is that the expression 1/2/3, when there are no parentheses, is defined to be (1/2)/3, which is different from the expression 1/(2/3), and that is why you got the wrong answer in your post. The second is why "the denominator of the denominator goes to the numerator", and it is not something you need to memorize, instead you can do the following
$$\dfrac1{\left(\frac23\right)} = \dfrac1{\left(\frac23\right)}\cdot 1 = \dfrac1{\left(\frac23\right)}\cdot \dfrac33 = \dfrac{1\cdot 3}{\left(\frac23\right)\cdot 3} = \dfrac{3}{\frac{2}3\cdot 3} = \dfrac32.$$
That is, if the denominator in the denominator is 3, just multiply the entire fraction by $\frac33$, and this will cancel the denominator in the denominator.
Best Answer
In the fraction $$\frac{p}{q}$$ where $q\ne 0$
$p$ is called Numerator and $q$ is called denominator