Let $K$ be a field. A vector space over $K$ is a set $V$, together with an operation $+\colon V\times V\to V$, and a function $\cdot\colon K\times V\to V$ (or alternatively, a family of functions $\lambda_c\colon V\to V$, indexed by elements $c\in K$) subject to certain identities and conditions.
The set $V$ does not normally contain $K$, so it makes no sense to talk about the vector space $V$ as being "closed under scalar multiplication". In addition, there is in general no "product" of vectors, so it does not make sense to talk about "vector multiplication."
Even in the case where we have a cross product of vectors (e.g., $\mathbb{R}^3$), this operation does not make $V$ into a group (the product is not associative).
Scalar multiplication is not an operation; an operation is always a function from a cartesian power of a set to the set. However, you can use currying to view the scalar multiplication as a family of unary operations on $V$, as indicated above, where for each $c\in K$ and each $\mathbf{v}\in V$, we define $\lambda_c(\mathbf{v})=c\mathbf{v}$.
(Added. Once you take into account the vector space axioms that relate to the scalar multiplication, it turns out that the scalar multiplication $K\times V\to V$ is actually an action of the field $K$ on the additive group $(V,+)$. But to truly call it an "action", you need the function to satisfy certain properties; just having a function $K\times V\to V$ does not make it an action, whereas any function $S\times S\to S$ is a binary operation on $S$, regardless of its properties.)
So pretty much all of your questions are based on misstatements of fact. Seems hard to answer them accurately.
Now, there is a situation in which some of what you say might make sense: if $F$ and $K$ are fields, and $F\subseteq K$, then we can view $K$ as a vector space over $F$ by "forgetting" about multiplication inside of $K$ when neither factor is in $F$. But here you really are going the other way: you already have a field, and you are obtaining a vector space by restricting the multiplication $\cdot\colon K\times K\to K$ to $F\times K$.
There is another situation in which you may have a product between "vectors": when you have an algebra. If $F$ is a field, a $K$ algebra is a ring with (a copy of) $F$ in the center of $K$. But again, what you have is a richer structure that, by "forgetting" part of the structure, yields a vector space. Much like you can have a ring and, by forgetting the product, obtain a group.
A concise mathematical term to describe the relationship between the Euclidean space $X = \mathbb E^n$ and the real vector space $V = \mathbb R^n$ is to say that $X$ is a principal homogeneous space (or ''torsor'') for $V$.
This is a way of saying that they are definitely not the same objects, but they very much are related to each other.
In particular:
- The objects $\mathbb E^n$ and $\mathbb R^n$ are exactly the same as sets of elements -- they both correspond bijectively to $n$-tuples of real numbers.
However,
in the vector space $\mathbb R^n$ we are allowed to add any two vectors (using the ''tip to tail'' visualization),
whereas in Euclidean space $\mathbb E^n$
there is no natural way to describe the process of ''adding'' two points.
Instead, given two points $P,Q$ in $\mathbb E^n$ we can naturally define their difference $\vec{v} = P-Q$, which is a vector in $ \mathbb R^n$.
This vector tells us how to get from point $Q$ to $P$ in $\mathbb E^n$.
in $\mathbb R^n$ there is a special ''zero vector'' $\vec{0} = (0,\ldots,0)$ which satisfies the additive property $\vec{0} + \vec{v} = \vec{v}$ for any $\vec{v}\in \mathbb R^n$,
while in $\mathbb E^n$ there is no point that is somehow more special that the other ones -- i.e. the space is ''homogeneous'' meaning it looks the same around every point.
for a vector $\vec{v}\in \mathbb R^n$ we can compute its length (or ''magnitute'' or ''norm'' etc.) by the formula
$$|\vec{v}| = \sqrt{v_1^2 + \cdots + v_n^2}.$$
For a point $P\in \mathbb E^n$, it does not make sense to ask what its length or distance is. It only makes sense to ask the distance between two points $P,Q$.
There are many natural examples of torsors motivated by physics, discussed in this blog post of John Baez, as well as more rigorous definitions if you are interested.
Best Answer
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
The operations on a vector space $\mathbb{V}$ over a field $\mathbb{F}$ are
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 $\{1, i\}$.