Please help me understand what a vector space over a field is? So for a set F where addition and multiplication are closed and the axioms are satisfied, F is a field.
For V to be a vector space, vector addition and scalar multiplication must be closed as well right?
Now for V to be a vector space over F, does that mean that V must first of all be closed under vector addition and scalar multiplication. Then all the elements of V must be elements of F as well. Is that right? The vector space over F must be contained in or equal to F?
 A: 
So for a set F where addition and multiplication are closed and the
  axioms are satisfied, F is a field.

There should be a $1 \in \mathbb{F}$ and subtraction and division should be closed as well, but that probably is an implication from the closedness of addition and multiplication and the axioms.

For V to be a vector space, vector addition and scalar multiplication
  must be closed as well right?

That is true the axioms should say that $V$ is an additive group regarding the addition (AEDDA) and an commutative multiplicative group regarding the scalar multiplication (ANIC).

Now for V to be a vector space over F, does that mean that V must
  first of all be closed under vector addition and scalar
  multiplication. 

We had that. The results must be in $V$.

Then all the elements of V must be elements of F as
  well. 

The scalars are taken from $\mathbb{F}$ and $V$ must be compatible with it. An 1-dimensional Vector space is probably isomorph to $\mathbb{F}$, if not identical. 
$$
V = \mathbb{R}, v = (x) \in V \wedge x \in \mathbb{R}
$$
But for example 
$$
V = \mathbb{R}^2 = \{ \left. v = (x, y)^T \,\right|\, x, y \in \mathbb{R} \}
$$ 
seems different from the field $\mathbb{R}$ to me. 

Is that right? The vector space over F must be contained in or
  equal to F?

Not in general. If they were always identical it would be a field and the vector space definition would be superfluous.
The scalar multiplication is not a multiplication between vectors, but 
between a field element (scalar) and a vector giving a vector.
In general there is no vector multiplication giving a vector and allowing a division of vectors.
Or said differently: A field is a vector space but not every vector space is a field.
Sorry for my comment about coordinate spaces, I am too used to them. 
So a coordinate space is a vector space. But not every vector space is a coordinate space.
E.g.
$$
V = C^1 = \{ \left. f: \mathbb{R} \to \mathbb{R} \,\right|\, f \mbox{ is differentiable on } \mathbb{R} \}
$$
is a vector space but no coordinate space (at least no finite dimensional one). I can add differentiable functions and get a differentiable function. I can also take any scalar $\alpha$ from $\mathbb{R}$ and use that to scalar multiply some $f \in V$ and get a
$(\alpha f)\in V$.
A: Note three out of eight axioms of vector space are related to scalar fields.


*

*$0\vec v=\vec 0$, where $0$ is additive identity of field $F$

*$1\vec v=\vec v$, where $1$ is multiplicative identity of field $F$

*$s(t\vec v)=(st)\vec v$, where $s,t\in F$ requires multipilcation closure $st\in F$


Generally speaking, $V$ and $F$ don't have set relation. Field $F$ provides a group action on $V$, called scalar multiplication $f:F\times V\to V$ with $(c,\vec v)\mapsto c\vec v$ satisfying axioms of vector space.
A: The abstract approach in mathematics should be motivated by concrete examples but should not be grounded in it. It should stand alone.
Let $V=\{x,y\}$ and define a binary operation $+$ on it satisfying the necessary properties. This will work if you let $x+y=y+x=y$ and $x+x=y+y=x$.
Next take a field $F=\{a,b\}$ where $a$ and $b$ are the additive and multiplicative identity respectively. Denote scalar multiplication by a $\odot$ and define it as follows: $a\odot x = a\odot y=b\odot x=x$ and $b\odot y=y$.
This algebraic structure qualifies as a vector space $V$ over the field $F$.
It is your own definition of scalar multiplication which is forcing the result to be in $V$. Otherwise $V$ and $F$ are two different sets. We may have $V=\{x=\mbox{apple},y=\mbox{banana}\}$ and $f=\{a=\mbox{giraffe},b=\mbox{tiger}\}$.
