How is a field different from a group? In my notes, I have the following definition:
"A field $(F,*)$ consists of set $F$ combined with a binary operation $+$ or $\cdot$, where


*

*The operation $+$ turns $F$ into an additive group with identity element $\underline{0}$.

*For all $a,b \in F$ we also have $ab \in F$ and $a \cdot b$ is commutative. It turns $F^{*}=\{x \in F | x \neq 0 \}$ into a multiplicative (abelian) group.


*

*$\cdot$ is distributive"



This is very similar to the definition of a group $(G,*)$. The only difference I can see in these definitions is that the operator $*$ used by a group is not restricted to just $+$ or $\cdot$.
I'm currently trying to understand fields in the context of vector spaces. Could someone also explain what is meant by "a vector space over $F$" (where $F$ is a field)? Does it mean the collection of vectors who's elements are all contained in $F$?
EDIT: I've just asked a professor a similar question and have been told that there are in fact 2 definitions of a field. The first is the definition being discussed in this thread, which is the "group theory definition". Similarly, there is the "vector calculus definition" which simply states that a field is a "function of position". 
Unfortunately, this has just confused me further, but at least I understand the reason behind my confusion.
 A: The first thing you should bear in mind is that fields have two operations, whereas groups only have one. (In your notes, the word "or" is misleading, as it should really be "and"). These two operations are more often than not called "addition" and "multiplication", so they are customarily denoted by $"+"$ and "$\cdot$", because the most natural examples of fields -- as the rationals and the reals, have precisely these two operations. 
A vector space over a field is yet another algebraic structure. It is similar to a group, in the sense that you can add together its elements, but it differs from a group inasmuch as you can multiply its elements by scalars. These scalars come from the underlying field $F$. So the vectors are not contained in the underlying field $F$. Vectors are not scalars.
A good way to internalize these concepts is to have a precise definition of a field and a vector space, and then check -- in full detail -- how these definitions are implemented in particular examples.
A: The binary operations for a field are not restricted to just $+$ or $\cdot$. Like $*$, the symbols $+$ and $\cdot $ are just generic symbols for arbitrary binary operations. However, a field needs two operations (i.e., the definition should read "$+$ and $\cdot$" instead of "$+$ or $\cdot$", and accordingly the tuple should be $(F,+,\cdot)$ instead of $(F,*)$). But to repeat: You might as well speak of a field $(F,*,\#)$ with two operations $*$ and $\#$, it need not be $+$ and $\cdot$. The use of the common symbols $+$ and $\cdot$ is motivated by the improved mnemonic. Or would you recall after a while which of $(a*b)\#c=(a\#c)*(b\#c)$ or $(a\#b)*c=(a*c)\#(b*c)$ holds in $(F,*,\#)$?
Of course there are relations to group theory:


*

*$(F,+)$ is a group (even an abelian group)

*$(F\setminus\{0\},\cdot)$, where $0$ is the neutral element of $+$, is a group (again, abelian)


And of course these are intertwined by the distributive property.
Note however that it is not possible to turn an arbitrary abelian group $(F,+)$ into a field $(F,+,\cdot)$ be finding a suitable operation $\cdot$.
A vector space over a field $(F,+,\cdot)$ is an abelian group $(V,\color{red}+)$ together with an action of $F$ on $V$, i.e., a ring homomorphism $\color{red}\cdot\colon F\to\operatorname{End}(V)$ (or spelled out a map $F\times V\to V$ such that  $(a+b)\color{red}\cdot v= a\color{red}\cdot v\color{red}+b\color{red}\cdot v$, $(a\cdot b)\color{red}\cdot v=a\color{red}\cdot(b\color{red}\cdot v)$, $a\color{red}\cdot(v\color{red}+ w)=a\color{red}\cdot v \color{red}+a \color{red}\cdot w$ and $1\color{red}\cdot v=v$) - colours only used to distinguish different operations using the same symbol for mnemonic reasons. So once again we see how much groups play a role, but also that it is necessary to specify the field when talking about vector spaces. For example, $\Bbb C$ is a two-dimensional vector space over $\Bbb R$, or a one-dimensional vector space over $\Bbb C$, or an infinite-dimensional vector space over $\Bbb Q$.
A: Uniquesolution gives a good answer. The answer to your second question "is a vector space over $F$ just a collection of vectors with coordinates in $F$?" is yes.
A particular vector space has dimension, which is just the number of different entries in a (column) vector drawn from the space. Two columns of different length have to live in different vector spaces. But the columns are allowed to be infinite.
A pure vector space doesn't know whether the coordinates of its vectors are columns, rows, matrices or whatever: a vector space of $n$-by-$m$ matrices is exactly the same as (isomorphic to) a vector space of column vectors of length $mn$. The difference between row and column is only visible when you want to define an operation between two vector spaces (like matrix multiplication, say) or some other structure. 
A: Your definition should read "combindend with a binary operation $+$ and $\cdot$", also it should better say that $+$ turns $F$ into an abelian group. Thereby the difference become more apparent:


*

*The set with the operator $+$ is not only a group, it's abelian (or commutative)

*The set (excluding the zero) with the operator $\cdot$ is also not only a group, it's abelian too.

*The distributive law holds.

