How to tell whether a vector of random variables has a single variable as input, and when it has a vector of variables as input? Let $X_1,...,X_n$ be random variables.
Then the formula
$$\mathbb{P}(\pmatrix{X_1\\...\\X_n} = A)$$
can have two meanings, dependent on how $\mathbb P $ is defined:
It can mean either $\mathbb{P}\left(\left\{w\in\Omega\mid \pmatrix{X_1(w)\\...\\X_n(w)} = z\right\}\right)$ , if we define our probability space as $(\Omega,\mathcal{F},\mathbb{P})$
or $\mathbb{P}\left(\left\{\pmatrix{w_1\\...\\w_n} \in \times_{i=1}^n \Omega_i \mid \pmatrix{X_1\\...\\X_n}\pmatrix{w_1\\...\\w_n}  = z\right\}\right)$
, if we define our probability space as $(\Omega^n,\mathcal{F},\mathbb{P})$.
However, defining the probability space is often skipped, and the domain of the $X_i$ isn't always denoted either.
If I happen to stumble upon such a case where I can't somehow deduce which is correct, is there a difference between the two cases, and which should I assume it is, then?
 A: Q.1. Is there a difference between the two cases?
A.1 Yes, I discern a difference between the two cases. 
Q.2 If I can't deduce which set of assumptions to use, which set of assumptions should I use?
A.2 Based on the law of parsimony, unless indicated otherwise, you should first assume the case that requires less words.  Notice that $\Omega$ is more parsimonious than $\Omega^n$. Therefore, first go with $\Omega$.
Q.3 Can you give me a more precise answer?
A.3 I think the way to treat this would be to consider something like the partition function.  You can use this process.  
By $m$, I denote a least least upper bound of $n$. For  $n \leq m$, the entropy will be 
$$\sigma^{(n)}_{X} = \sum_{i\in n} p_i   \log(p_i)$$
So you obtain a sequence
$$\left\{\sigma^{(1)}_{X}, \sigma^{(2)}_{X}, \ldots, \sigma^{(m)}_{X}\right\}$$
Lets get a sense of things, by looking at the entropys' expected values under equi-partition. The entropys' expected values under equipartition  are
$$\left\{\left< \sigma^{(1)}_{X}\right>, \left<\sigma^{(2)}_{X}\right>, \ldots, \left<\sigma^{(m)}_{X}\right>\right\}= \left\{-\log(1), -\log(2), \ldots, -\log(m)\right\}.$$
Observe that the sequence diverges.
One ask, which one of these gives the best and most correct indicator? Notice for $q<p$, that expected value of $\left<\sigma^{(q)}\right> > \left<\sigma^{(p)}\right>$. Also ponder the following: in getting $\left<\sigma^{(p)}\right>$, I had to average all possibilities---including for all possibilities when there were only $q$ non-zero random variables. 
This insinuates that you should look at your expected  results for all $n\in \mathbb{Z}^+$, and also in the limit that $n$ goes to infinity. My guess is that though the partition function diverges with increasing $n$, the thing that you would actually care to and be able to observe would converge.
A: So, in general it is: When you deal with a function $f:\mathbb R \to \mathbb R^n$, then you can write it as a $f=(f_1,...,f_n)$ where $f_k:\mathbb R \to \mathbb R$ is a single variable real valued function for $k \in \{1,..,n\}$. In that case $f(t) = (f_1(t),...,f_n(t))$ and you probably won't have any doubt, it is the way that we should look at it. The same goes if our function have a domain in more abstract space, that is, Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $X:\Omega \to \mathbb R^n$ be random variable. It can be shown, that writing $X=(X_1,...,X_n)$, then $X_k : \Omega \to \mathbb R$ is a random variable for any $k \in \{1,...,n\}$. In that case $X(\omega) = (X_1(\omega),...,X_n(\omega))$ and the proper way to look at it is $\mathbb P(X \in A) = \mathbb P (\{\omega \in \Omega : (X_1(\omega),...,X_n(\omega)) \in A \})$, since $\mathbb P$ is a measure on $(\Omega,\mathcal F)$. So when you have just a random vector, then you shouldn't think that every coordinate takes different argument. Every coordinate should be a function from whole space (that is $\Omega$) no matter what $\Omega$ looks like.
Okay, it was the case, when we had a $\mathbb R^n$ valued random variable and made it into the vector of $\mathbb R$ valued random variables. However, there are other possibilities. Instead of having vector and looking at it coordinates, we can have a lot of random variables and form a new vector. However, it isn't as simple as it might look: If you have probability spaces $(\Omega_1,\mathcal F_1,\mathbb P_1),...,(\Omega_n,\mathcal F_n, \mathbb P_n)$ and defined real valued random variables on them: $X_k : \Omega_k \to \mathbb R^n$, you can define new  set, call it $\Omega$ which would be defined as $\Omega = \Omega_1 \times ... \times \Omega_n$ (so every $\omega \in \Omega$ is of the form $(\omega_1,...,\omega_n)$ where $\omega_k \in \Omega_k$), and (call it for now function, since we didn't specified sigma fields) function $X:\Omega \to \mathbb R^n$ given by $X(\omega) = (X_1(\omega_1),...,X_n(\omega_n))$. And that is obviously another proper way to look at it, when considering $X$ just as a function, but it is more subtle when considering it as a random, measurable function! As for the case with measurability, you can always define new sigma field as $\mathcal F = \mathcal F_1 \otimes ... \otimes \mathcal F_n =: \sigma( A_1 \times ... \times A_n : A_k \in \mathcal F_k , k \in \{1,..,n\})$ (loosely speeking, you just take any $"$rectangle$"$ of the base elements of every $\mathcal F_k$ which is in form $A_1 \times ... \times A_n$ where $A_k \in \mathcal F_k$ and close it under operationts that are needed to form a $\sigma-$field.) Now, the problem with measuring (so calculating probability) isn't as easy, it requires the concept of Product measure (which you can google). Again, loosely speaking, it defines a measure $\mathbb P$ on our new measurable space $(\Omega,\mathcal F)$ to be $\mathbb P(A_1 \times ... \times A_n) = \mathbb P_1(A_1) \cdot ... \cdot \mathbb P_n(A_n)$ for every rectangle $A_1 \times ... \times A_n$ (in case of your random variable it would mean that $\mathbb P(X \in B_1 \times ... \times B_n) = \mathbb P(\{\omega \in \Omega : X(\omega) \in B_1 \times ... \times B_n\}) = \mathbb P_1(\{\omega_1 \in \Omega_1 : X_1(\omega_1) \in B_1 \}) \cdot ... \cdot \mathbb P_n(\{\omega_n \in \Omega_n : X_n(\omega_n) \in B_n \}) = \mathbb P_1(X_1 \in B_1)...\mathbb P_n(X_n \in B_n)$ (note that we're calculating probability for every $X_k$ on the other space and with respect to different probability measure, since variables $X_1,...,X_n$ arn't defined on $\Omega$ but on $\Omega_1,...,\Omega_n$ respectivelly). It (again google product measure) can be shown that when the space is $\sigma-$finite then it is uniquelly determined.
It was a long story, but what is important, that when defining it as a product space, then every coordinate would be INDEPENDENT! (note the measure $\mathbb P$ is defined so, since $\mathbb P_k(X_k \in B_k) = \mathbb P( X_k \in B_k, X_1,...,X_{k-1},X_{k+1},...,X_n \in \mathbb R )$) That means by considering a random vector in the way that coordinates have different arguments, it would make them independent, and we're loosing a lot of possible random vector, where the coordinates need not be independent.
So the lesson should be, in general, when dealing with random vector $X$ on $(\Omega,\mathcal F,\mathbb P)$ you can write it as $X=(X_1,...,X_n)$ where every $X_k : \Omega \to \mathbb R$ is a random variable and $X(\omega) = (X_1(\omega),...,X_n(\omega))$ since it is just defined this way. However, when you know that $X_1,...,X_n$ are independent, then you can REDEFINE (it won't be exactly the same random variable, just similar, with the same distribution) it as in the product space, taking every $\Omega_k = \Omega$ and $X:\Omega^n \to \mathbb R^n$ will be given as $X((\omega_1,..,\omega_n)) = (X_1(\omega_1),...,X_n(\omega_n))$. That sometimes is an useful approach when one is interested in things concerning only the distribution of $X$.
