Thanks to the stimulating discussion with @Didier, I've clarified something for myself. From the technical standpoint, we have $n$ random variables in the option 1, and $n$ numbers in the option 2. The problem with this may be illustrated by the following example. Consider 3 different people producing their own random samples of size 5 by throwing a die. Here is what they get:
1 person: 1, 3, 1, 4, 2 $\;\;\;\rightarrow$ X_1(\omega'), X_2(\omega'), ..., X_5(\omega')
2 person: 2, 2, 1, 6, 3 $\;\;\;\rightarrow$ X_1(\omega''), X_2(\omega''), ..., X_5(\omega'')
3 person: 1, 4, 2, 1, 3 $\;\;\;\rightarrow$ X_1(\omega'''), X_2(\omega'''), ..., X_5(\omega''')
On the right side, I used option 1 to code these outcomes. How to code them using option 2? We can try this:
1 person: 1, 3, 1, 4, 2 $\;\;\;\rightarrow$ $X(t_1), X(t_2), ..., X(t_5)$
2 person: 2, 2, 1, 6, 3 $\;\;\;\rightarrow$ $X(t_1), X(t_2), ..., X(t_5)$
3 person: 1, 4, 2, 1, 3 $\;\;\;\rightarrow$ $X(t_1), X(t_2), ..., X(t_5)$
($t$ is for "trial number"). Well, this clearly doesn't work. What about this:
1 person: 1, 3, 1, 4, 2 $\;\;\;\rightarrow$ $X(\omega_1), X(\omega_3), ..., X(\omega_2)$
2 person: 2, 2, 1, 6, 3 $\;\;\;\rightarrow$ $X(\omega_2), X(\omega_2), ..., X(\omega_3)$
3 person: 1, 4, 2, 1, 3 $\;\;\;\rightarrow$ $X(\omega_1), X(\omega_4), ..., X(\omega_3)$
This seems to work, but how to write it in a general manner?
person ?: ?, ?, ?, ?, ? $\;\;\;\rightarrow$ $X(?), X(?), X(?), X(?), X(?)$
I think it is this point at which we arrive to an appropriate general notation specified in the option 1.