What makes a vector an object with both magnitude and direction? According to my understanding, A vector is an element of a set called the vector space which satisfies a list of axioms like : closure under vector addition, closure under scalar multiplication, associativity, commutativity, distributivity, the existence of : a zero vector(additive identity), multiplicative identity , additive inverse and so on. 
They're quite useful in physics for at least two reasons:
1)a vector is a quantity that has both magnitude and direction, and such a quantity pops up a lot in physics.
2)a vector is invariant under co-ordinate rotation and translation. Now this is pretty important because: if a physical law can be described by vector equation (e.g. Newton's second law) then this law is invariant under co-ordinate rotation and translation, a property that every physical law should satisfy.
My question is : How does an element in vector space(i.e. a vector) which satisfies the aforementioned list of axioms imply that this element(this vector) is a quantity that has both magnitude and direction?
In addition to that, How does satisfying the aforementioned axioms make this element(this vector) invariant under rotation and translation of coordinates?
 A: Elements of a real vectorspace certainly have direction, but they don't really have a magnitude. Well actually, they... kind-of have a magnitude. But for a proper magnitude, you need further structure, such as a norm or inner product. Let me explain.
Vector Spaces.
Suppose $V$ is a real vectorspace.

Definition 0. Given a vectors $x,y \in V$, we say that $x$ and $y$
  have the same direction iff:
  
  
*
  
*there exists $r \in \mathbb{R}_{\geq 0}$ such that $x = ry,$ and
  
*there exists $r \in \mathbb{R}_{\geq 0}$ such that $y = rx$.
  
  
  (The $r$'s don't have to be the same.)

This induces an equivalence relation on $V$, so we get a partitioning of $V$ into cells. Each cell is an open ray, so long as we regard $\{0\}$ as an open ray. You may wish to exclude $\{0\}$ from its privileged position as a ray, in which case you should only deal with non-zero vectors; that is, you need to be dealing with $V \setminus \{0\}$ rather than $V$.
Irrespective of which conventions are used, we can make sense of direction using these ideas:

Definition 1. The direction of $x \in V$ is the unique open ray $R \subseteq V$ such that $x \in R$.

Notice that the equivalence relation of having the same direction is preserved under scalar multiplication; what I mean is that if $v$ and $w$ have the same direction, then $av$ and $aw$ have the same direction, for any $a \in \mathbb{R}$. Geometrically, this means that if we scale a ray, we'll end up with a subset of another ray.
As for magnitude; well, if you choose a ray $R \subseteq V$, then we can partially order $R$ as follows. Given $x,y \in R$, we define that $x \geq y$ iff $x = ry$ for some $r \in \mathbb{R}_{\geq 1}$. So some vectors along this ray are longer than others, hence magnitude.
Inner Product Spaces.
Actually, this isn't the whole story. The problem with vector spaces is that if $x$ and $y$ don't belong to the same ray (nor to the the "negatives" of each others rays), then there's no way of comparing the magnitudes of $x$ and $y$. We can't say which is longer! Now there are mathematical situations where this limitation is desirable, but physically, you probably don't want this. A related issue is that you can't really make sense of angles in a (mere) vector space; at least, not without some further structure.
For this reason, when physicists say "vector", what they usually mean is "element of a finite-dimensional inner-product space." This is a (finite-dimensional) vector space $V$ with further structure; in particular, it comes equipped with a function
$$\langle-,-\rangle : V \times V \rightarrow \mathbb{R}$$
that is required to satisfy certain axioms resembling the dot product. Especially important for us is that these axioms include a "non-negativity" condition:
$$\langle x,x\rangle \geq 0$$
Using this, we can define the magnitude of vectors as follows.

Definition 2. Suppose $V$ is a real inner product space. Then the norm (or "magnitude") of $x \in V$, denoted $\|x\|$, is defined a follows:
$$\|x\| = \langle x,x\rangle^{1/2}$$

This allows us to compare the magnitudes of vectors that don't live in the same ray; we simply define that $x \geq y$ means $\|x\| \geq \|y\|.$ When confined to a single ray, this agrees with our earlier definition! Be careful though, because the relation $\geq$ we just defined is only a preorder.
In fact, the inner product gives us more than just magnitudes; it also gives angles!

Definition 3. Suppose $V$ is a real inner product space. Then the angle between of $x,y \in V$, denoted $\mathrm{ang}(x,y)$, is defined a follows:
$$\mathrm{ang}(x,y) = \cos^{-1}\left(\frac{\langle x,y\rangle}{\|x\|\|y\|}\right)$$

It can be shown that vectors $x$ and $y$ have the same direction (in the sense described at the beginning of my post) iff the angle between them is $0$. In fact, you can modify the above definition so that it defines the angle between any two non-zero open rays. In this case, it turns out that two rays are equal iff the angle between them is $0$.
A: The word "vector" is used to mean different things in math.  Sometimes "vector" means "an ordered $n$-tuple of real numbers".  Sometimes it means an equivalence class of directed line segments (with respect to a certain equivalence relation).  Sometimes it means an element of a particular vector space.  Which definition is being used depends on context.  If a vector space does not have a norm defined on it, then indeed the elements of that vector space do not have magnitudes.  In that context the "magnitude" of a vector has not been defined.
