First, I understand that $\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X \mid Y]]$, but
how to prove that $$\mathbb{E}[\mathbb{E}[X \mid Y, Z] \mid Z] = \mathbb{E}[X \mid Z]?$$
Second, for $\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X \mid Y]]$, I knew an intuitive interpretation about "computing an average IQ of a population by first computing the average IQs of men and women respectively and then summing the two averages, with each weighted by the respective rate of men and women".
Could you come up with any similar interpretations of the second equation $$\mathbb{E}[\mathbb{E}[X \mid Y, Z] \mid Z] = \mathbb{E}[X \mid Z]?$$
Assuming everything is discrete, the conditional expectation is defined as $\mathbb{E}[X\mid Z]=\varphi(Z)$ where $$ \varphi(z)=\mathbb{E}[X\mid Z=z]=\sum_x x p_{X\mid Z}(x\mid z) $$ with $p_{X\mid Z}(x\mid z)=p_{X,Z}(x,z)/p_Z(z)$ for $p_Z(z)>0$ and zero elsewhere.
Show that $U:=\mathbb{E}[X\mid Z]$ defined as above is equivalent to requiring that $$ \mathbb{E}[U\mathbf{1}_{Z\in A}]=\mathbb{E}[X\mathbf{1}_{Z\in A}] $$ for all $A\subseteq\mathbb{R}$.
Using the equivalent definition of 1. we thus need to show that $$ \mathbb{E}[V\mathbf{1}_{Z\in A}]=\mathbb{E}[X\mathbf{1}_{Z\in A}] $$ for all $A\subseteq\mathbb{R}$ with $V=\mathbb{E}[X\mid Y,Z]$. But this is obvious by using the equivalent definition of $V=\mathbb{E}[X\mid Y,Z]$ since we know that $$ \mathbb{E}[V\mathbf{1}_{(Y,Z)\in C}]=\mathbb{E}[X\mathbf{1}_{(Y,Z)\in C}] $$ for all $C\subseteq\mathbb{R}^2$. Now simply let $C=\mathbb{R}\times A$.