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 $\{1, i\}$.