Confusions about definition of the sum of two or several random variables Description of the Questions:
I have some trouble understanding the definition of the sum of two random variables and hope someone could help me to get it straight. And my main confusions could be divided into the following questions:
First of all, the usual definition of the sum of two random variables is given by:

Definition: Sum of two random variables
Let $X$ and $Y$ be random variables defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, then define a new random variable $Z(\omega)$ on the same probability space as: $Z(\omega) = X(\omega)+Y(\omega). \quad \forall \omega \in \Omega $
Notice: It is necessary to verify that this definition is well defined. i.e. we need to verify that $Z(\omega)$ is indeed a random variable. But we could simply leave it aside since this proof is not required in my question.

Now, if we consider the following simple example:
Suppose we have a given probability space $(\Omega, \mathcal{F}, \mathbb{P})$, where $\Omega = \{\omega_1, \omega_2, \omega_3, \omega_4\}$ & $\mathcal{F} = \mathcal{P}(\Omega)$ and a proper probability measure $\mathbb{P}$. Also, suppose we have two same discrete random variable $X_1(\omega)$ and $X_2(\omega)$ defined as follow:
$$X_1(\omega_i) = X_2(\omega_i) = i \quad \forall \omega_i \in \Omega $$
Finally, we could define the sum of these two variables as:
$$Z(\omega) = X_1(\omega) + X_2(\omega) \quad \forall \omega \in \Omega$$
But, According to this definition, when $\omega_i$ goes through all the possible value in $\Omega \quad$ i.e.  $i \in \{ 1,2,3,4\}$, $Z(\omega) \in \{ 2,4,6,8\}$ 

I'm fully aware that this result is due to the understanding that restricting $\omega_i$ to be the same event in both $X_1$ and $X_2$
Question 1:  From the angle of strictly following the definition, Am I somehow wrong or not in understanding the definition?
But when calculating of probability measure $\mathbb{P}(Z(\omega)=z) = \displaystyle{\sum_{x_{1}}p_{X_1,X_2}(x_1,z-x_2)}$, we could see that $\mathbb{P}(Z(\omega)=3)$ exists and equals to $ \displaystyle\sum_{x_{1}}p_{X_1,X_2}(x_1,3-x_2) = p_{X_1,X_2}(1,2) + p_{X_1,X_2}(2,1)$
I also know that if we allow that $\omega_i$ to be different in $X_1$ and $X_2$, then we should probably consider a 2-dimensional space $\Omega \times \Omega$, i.e. $Z(\delta = (\omega_1,\omega_2)) = X_1(\omega_1) + X_2(\omega_2) \quad \forall \omega_1, \omega_2 \in \Omega, \delta \in \Omega \times \Omega$
Question 2: this definition might work, and also conform to probability calculation, but here $\delta \notin \Omega$, which means $Z(\delta)$ is defined in a different probability space, and thus we need to re-evaluate the well-definedness of such definition, is this idea correct?
 A: By definition of a random variable it maps to $\mathbb{R}$ so that is where $Z(\omega)$ lands. And yes, when you look at the sum of two random variables you look at a new variable whose domain is a product set of the original $\Omega$. Measurability is defined in a broad way relating image-subsets to domain-subsets. There is no need at all to re-evaluate the well-definedness but rather simply to prove measurability of $Z$. Which you can do like it is done in this post:
Proving that sum of two measurable functions is measurable.
Edit:
In general measurability is defined like this:
https://en.wikipedia.org/wiki/Measurable_function
In the case of random variables the definition is that X needs to be measurable w.r.t. $(\mathbb{R},\mathfrak{B})$ where $\mathfrak{B}$ is the Borel-$\sigma$-Algebra. This means that for any $A \in \mathfrak{B}$ we have that
$X^{-1}(A) \in \mathfrak{F}$
where $\mathfrak{F}$ is the $\sigma$-Algebra of your probability space $(\Omega, \mathfrak{F}, P)$. A prime reason why we demand this property for random variables is that it allows us to use a Pushforward-Measure on $\mathbb{R}$:
https://en.wikipedia.org/wiki/Pushforward_measure
That means we can assign probabilities to random events in $\mathbb{R}$  (e.g. the sum of two throws of a dice) using the probability measure we already have on $\Omega$.
