So, let’s address this question first: ''What is the basic difference between an algebraic variable and a Random Variable?''
A random variable $X$ is not at all an algebraic variable. Formally, it is defined as a function from a probability space $\Omega$ to $\mathbb R$.
OK... What that really means is that you perform random experiments (eg throwing a dice, choosing a random human), and you make measures on these experiments (eg number on dice upper face, height, sex, cholesterol level of the human). The set $\Omega$ is the set of all possible experiments. On a particular experiment $\omega\in\Omega$, you make a measure $X(\omega)$: that’s why formally it is a function from $\Omega$ to $\mathbb R$.
Now in general we totally forget about $\Omega$. The random variables are defined in term of their probability law. In the case of a fair dice, you just say
- $\mathbb P(X = k) = {1\over 6}$ for $k=1,\dots,6$ (the probability of $X$ equal to $k$ is 1/6 for $k$ from 1 to 6),
instead of
- $\mathbb P\left( \bigl\{ \omega \in \Omega \ : \ X(\omega) = k\bigr\}\right)$ (the set of dice throws on which the measure $X$ — upper face — is $k$ has probability 1/6)...
It’s simpler. You can even totally avoid bothering the students with $\Omega$.
I hope this sheds some kind of light.
Now what this guy meantmeans by $X + X \ne 2X$ isn’t that the sum of such a measure with itself is not twice this measure — unfortunately, it is what he writes. What he means is that the sum of two such measures, performed on different experiments, has not the same law than twice a measure. This could be written as $X_1 \sim X_2 \not\Rightarrow X_1 + X_2 \sim 2X_1$ (the fact that $X_1$ and $X_2$ have the same distribution does not imply that $X_1 + X_2$ has the same distribution hasas $2 X_1$).