Suppose that we have a stochastic process of the form $X_t=t\cdot Z$ where $Z\sim N(0,1)$. Anyone who has ever done a course in stochastic processes will be able to tell you that the realizations of this process will only differ due to the value of $t$ increasing. For example, we will always have $2X_1=X_2$. Similarly, we can see that the distribution $Z$ is only ever sampled once in any realization of the entire stochastic process.
My question is as follows - why is $Z$ only sampled once? I can understand the practical reasons for it, but what specific definition is responsible for this? It's certainly not part of any definitions of random variable or stochastic process that I've ever seen.
Everything depends on the definition of a stochastic process, in particular, the corresponding probability space makes a big contribution; it is not often even mentioned as people find it either obvious (after you do a course in measure theory) or they rely on their intuition (less rigorous approach often without measure theory background).
I will try to explain it on a very simple example that is a bit similar to the one you mentioned, I will just use discrete time processes and also instead a normal random variable I will use a Bernoulli one.
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be as follows $\Omega = \{Heads, Tails\}$, $\mathcal{F} = 2^{\Omega}$ and $\mathbb{P} \colon \mathcal{F} \rightarrow [0, \infty)$ be defined as $\mathbb{P}(\{Heads\}) = \mathbb{P}(\{Tails\}) = 0.5$. Define a random variable $Z \colon \Omega \rightarrow \{-1, 1\}$ such that $Z(Heads) = 1$, $Z(Tails)= -1$. Clearly, $Z$ is a Bernoulli (or Radomacher depending on nomenclature) random variable, in particular its distribution is given by $\mathbb{P}(Z=1) = \mathbb{P}(Z=-1) =0.5$. Let us construct a discrete stochastic process $X=(X_n)_{n \in \mathbb{N}}$ such that $X_n = n Z$, for each $n \in \mathbb{N}$. For any $n$, $nZ$ is random variable on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$. We have clearly only two sample paths (realisations) of this process that is, for $\omega = Heads$ we have $\{n \colon n \in \mathbb{N}\}$ and for $\omega = Tails$ we have $\{-n \ \colon n \in \mathbb{N}\}$.
Now, let us consider a different process, and let $Y$ be an independent copy of $Z$, that is $Y$ has the same distribution as $Z$ and they are both independent. We define a stochastic process $X=(X_n)_{n \in \mathbb{N}}$ with the probability space will be $( \Omega \times \Omega, 2^{\Omega \times \Omega}, \mathbb{P} \times \mathbb{P})$ as follows $$X_n = \begin{cases}nY &\mbox{ if } n \mbox{ is odd} \\ nZ & \mbox{ if } n \mbox{ is even } \end{cases}.$$ In particular, for the above definition to make sense, for any $\omega = (\omega_1, \omega_2) \in \Omega \times \Omega$ we identify $Z(\omega) \equiv Z(\omega_1)$ and $Y(\omega) \equiv Y(\omega_2)$; so that each $X_n$ are defined on the same probability space.
Therefore, we have four paths in total, for $\omega = (Heads, Heads)$ we have $\{ n \colon \mbox{ odd}\} \cup \{ n \colon n \mbox{ even}\} = \mathbb{N}$, for $\omega = (Tails, Heads) $ we have $\{ -n \colon n \mbox{ even} \} \cup \{ n \colon n \mbox{ odd}\}$ and so on for the other choices from $\Omega \times \Omega$.
So your sampling depends on the definition of a stochastic process itself; what is the underlying probability space, and how you identify your random variables. I hope that I was clear enough.