Firstly, a vector space need not be a copy of $\mathbb R^n$, it can be an arbitrary set with addition and scalar multiplication defined according to the vector space axioms.
Secondly, you can always formally distinguish between two copies of a set $A$ by considering $A\times\{0\}$ and $A\times\{1\}$. So you could say that points are members of $\mathbb R^n\times \{0\}$ and vectors memebers of $\mathbb R^n\times \{1\}$. And in fact some mathemeticians, if pressed, will take that as their formal justification.
However, there are some more elegant ways to define these things, though they take a bit of machinery to do. You can define what's called an affine space. A traditional way of doing this (which they do at the Wikipedia link as of this writing) is to define an affine space $(\mathbb A, \mathbb T,\tau)$ to be a set $\mathbb A$ together with a vector space of translations $\mathbb T$ and an action $\tau$ of $\mathbb T$ (viewed as an additive group) on $\mathbb A$ that is transitive and free.
You can also equivalently define an affine space as a set $\mathbb A$ with formal affine combinations. For this a full treatment is a little involved, but the basic idea is you consider the free vector space of $\mathcal F(\mathbb A)$ of formal linear combinations of members of $\mathbb A$, take the subset of affine combinations $\mathcal A(\mathbb A)$ (linear combinations whose scalars sum to $1$), and define a mapping from $\mathcal A(\mathbb A)$ to $\mathbb A$ that satisfies reasonable properties.
This allows you to make formal sense of $\lambda_1 a_1\cdots + \lambda_n a_n\in\mathbb A$ whenever $\sum_{k=1}^{n}\lambda_k=1$ and each $a_k\in\mathbb A$. Then you define vectors to be equivalence classes $\vec{ab}$ of pairs $(a,b)$, where the equivalence relation $\sim$ is given by $(a,b)\sim(c,d)$ if and only if $a + d - c = b$. You can show that this set of vectors has a natural vector space structure given by $\vec{ab}+\vec{cd}=\overrightarrow{a(b+d-c)}$ and $\lambda\vec{ab}=\overrightarrow{a(a+ \lambda b - \lambda a)}$.
With this interpretation, if you want you can let your affine space (i.e., your point set) be $\mathbb R^n$ with affine combinations defined in the obvious way, and then your vectors are just $\vec{pq}$ with $p,q\in\mathbb R^n$.