Firstly, how do I interpret $\mathbb{E}[g(X)]$. I understand $\mathbb{E}[X]$ is like the most likely outcome of a set of experiments (loosely speaking at least - not really a very maths person)? But $\mathbb{E}[g(X)]$? How do I relate to a real life example?
Then, I know $E[g(X)] = \sum_X g(x) f(x)$ but how do I find $\mathbb E[Y g(X)]$
Why is
$$\mathbb{E} [Y g(X)] = \sum_{(x,y) \in Z} y \, g(x) f(x, y)$$
How do I get this?
It looks like we have a random variable $X$ with probability mass function $f$. We also have some function $g$ that maps the range of $X$ to some other real number.
Informally, if $X=x$, we replace it by $g(x)$. Then we take the expected value.
An simple example where this occurs: Suppose we had a fair coin with one side labeled $0$ and one side labeled $1$. Then the expected value of a coin toss would be $$\mathbb{E}[X]=\sum_{x \in \mathrm{range}(X)} x f(x)=0 \times \tfrac{1}{2}+1 \times \tfrac{1}{2}=0.5.$$
Now suppose we took the coin, and replaced the labels $0 \mapsto 1$ and $1 \mapsto 2$, that is, we replace $x$ with $g(x)=x+1$. Now the expected value is $$\mathbb{E}[g(X)]=\sum_{x \in \mathrm{range}(X)} g(x) f(x)=\sum_{x \in \mathrm{range}(X)} (x+1) f(x)=1 \times \tfrac{1}{2}+2 \times \tfrac{1}{2}=1.5.$$
In the second part of the question, we have an additional random variable $Y$. Or, equivalently, we have a random variable $(X,Y)$ with probability mass function $f(x,y)$.
So $$\mathbb{E}[Y g(X)]=\sum_{(x,y) \in \mathrm{range}((X,Y))} y g(x) f(x,y).$$
Informally, if $(X,Y)=(x,y)$, we replace it by $y g(x)$. Then we take the expected value.