Recall that if a vector space has a finite basis it is said to be finite dimensional, and the dimension is defined to be the number of vectors that make up this basis. Basis are (possibly finite) sets of vectors that span the vector space and are linearly independent. One can prove that every vector in said vector space can be written in one and only one way as a linear combination of these basis vectors. Say $V$ is a $K$ vector space with basis $B=\{v_1,\ldots,v_n\}$. Then if we have $$v=\alpha_1v_1+\cdots+\alpha_nv_n$$
we write $(v)_B=(\alpha_1,\ldots,\alpha_n)$ and say $v$ has coordinates $(\alpha_1,\ldots,\alpha_n)$ in the basis $B$. This immediately gives a mapping $V\to F^n$ given by $$v\mapsto (v)_B$$
This is the same as mapping each basis vector $v_i$ to $$(0,0,\ldots,\underbrace{1}_i,\ldots,0)$$
which entirely determines the transformation.
Note that $0\mapsto (0,0,\ldots,0)$; that $(v+w)_B=(v)_B+(w)_B$ and $(\lambda v)_B=\lambda (v)_B$ so this is a linear transformation, which gives an isomorphism between $V$ and $F^n$. This means $V$ and $F^n$ are essentially the same as vector spaces, that is, "there is only one vector space of dimension $n$ over a field $F$ up to isomorphism."
It is true that vector spaces and fields both have operations we often call multiplication, but these operations are fundamentally different, and, like you say, we sometimes call the operation on vector spaces scalar multiplication for emphasis.
The operations on a field $\mathbb{F}$ are
- $+$: $\mathbb{F} \times \mathbb{F} \to \mathbb{F}$
- $\times$: $\mathbb{F} \times \mathbb{F} \to \mathbb{F}$
The operations on a vector space $\mathbb{V}$ over a field $\mathbb{F}$ are
- $+$: $\mathbb{V} \times \mathbb{V} \to \mathbb{V}$
- $\,\cdot\,$: $\mathbb{F} \times \mathbb{V} \to \mathbb{V}$
One of the field axioms says that any nonzero element $c \in \mathbb{F}$ has a multiplicative inverse, namely an element $c^{-1} \in \mathbb{F}$ such that $c \times c^{-1} = 1 = c^{-1} \times c$. There is no corresponding property among the vector space axioms.
It's an important example---and possibly the source of the confusion between these objects---that any field $\mathbb{F}$ is a vector space over itself, and in this special case the operations $\cdot$ and $\times$ coincide.
On the other hand, for any field $\mathbb{F}$, the Cartesian product $\mathbb{F}^n := \mathbb{F} \times \cdots \times \mathbb{F}$ has a natural vector space structure over $\mathbb{F}$, but for $n > 1$ it does not in general have a natural multiplication rule satisfying the field axioms, and hence does not have a natural field structure.
Remark As @hardmath points out in the below comments, one can often realize a finite-dimensional vector space $\mathbb{F}^n$ over a field $\mathbb{F}$ as a field in its own right if one makes additional choices. If $f$ is a polynomial irreducible over $\mathbb{F}$, say with $n := \deg f$, then we can form the set
$$\mathbb{F}[x] / \langle f(x) \rangle$$
over $\mathbb{F}$: This just means that we consider the vector space of polynomials with coefficients in $\mathbb{F}$ and declare two polynomials to be equivalent if their difference is some multiple of $f$. Now, polynomial addition and multiplication determine operations $+$ and $\times$ on this set, and it turns out that because $f$ is irreducible, these operations give the set the structure of a field. If we denote by $\alpha$ the image of $x$ under the map $\mathbb{F}[x] \to \mathbb{F}[x] / \langle f(x) \rangle$ (since we identify $f$ with $0$, we can think of $\alpha$ as a root of $f$), then by construction $\{1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}\}$ is a basis of (the underlying vector space of) $\mathbb{F}[x] / \langle f \rangle$; in particular, we can identify the span of $1$ with $\Bbb F$, which we may hence regard as a subfield of $\mathbb{F}[x] / \langle f(x) \rangle$; we thus call the latter a field extension of $\Bbb F$. In particular, this basis defines a vector space isomorphism
$$\mathbb{F}^n \to \mathbb{F}[x] / \langle f(x) \rangle, \qquad (p_0, \ldots, p_{n - 1}) \mapsto p_0 + p_1 \alpha + \ldots + p_{n - 1} \alpha^{n - 1}.$$ Since $\alpha$ depends on $f$, this isomorphism does depend on a choice of irreducible polynomial $f$ of degree $n$, so the field structure defined on $\mathbb{F}^n$ by declaring the vector space isomorphism to be a field isomorphism is not natural.
Example Taking $\Bbb F := \mathbb{R}$ and $f(x) := x^2 + 1 \in \mathbb{R}[x]$ gives a field
$$\mathbb{C} := \mathbb{R}[x] / \langle x^2 + 1 \rangle.$$
In this case, the image of $x$ under the canonical quotient map $\mathbb{R}[x] \to \mathbb{R}[x] / \langle x^2 + 1 \rangle$ is usually denoted $i$, and this field is exactly the complex numbers, which we have realized as a (real) vector space of dimension $2$ over $\mathbb{R}$ with basis $\{1, i\}$.
Best Answer
In a field you can multiply any two elements, as well as add them, while in a vector space you can add vectors, but you can only multiply by scalars; you can't multiply two vectors. (Multiplication in a field has to satisfy some axioms, but I think this is the essential point for your question.)
A field containing $\mathbb R$, or more generally a system of hypercomplex numbers containing $\mathbb R$, will in particular be a vector space (since you can add, and you can multiply by elements of $\mathbb R$, i.e. by scalars), but it will have extra structure. (E.g. you can multiply quaternions, and this is extra structure which is not at all obvious if you just know about $4$-vectors.)