Why isn't the scalar product of a covector and vector symmetric? In tensor math, how come the scalar product of a co-vector (co-variant vector) with contra-variant vector, as written between angle bracket separated by comma, $\langle x, a \rangle$, is not symmetric? 
 A: $\def\cV{{\cal V}}$
I don't know the book you quoted and am not interested in it. It's simply wrong, according universal use in mathematics, to call your $\langle x,a\rangle$ a scalar product.
Given a vector space $\cV$ on a field $K$, its dual space $\cV^*$ is defined as the set of linear functions 
$$x: \cV \to K,\quad a \in \cV \mapsto \langle x,a\rangle \in K$$ with linearity defined as usual.
This is possible for whatever vector space, with no other special structure required for it. It can easily be shown that $\cV^*$ can be given the structure of vector space on $K$. Elements of $\cV$ are called vectors, those of $\cV^*$ are called covectors. Of course it would be meaningless to write
$$\langle x,a\rangle = \langle a,x\rangle$$
as $a$ and $x$ belong to different spaces.
A scalar product is a structure of $\cV$ not always required or useful. When required it's defined as follows.
A scalar (inner) product on $\cV$ is a bilinear,  symmetric, non-degenerate mapping 
$$\cV \times \cV \to K,\quad 
(a,b) \in \cV \times \cV\ \mapsto\ a\cdot b \in K.$$
Here $(a,b)$ means the ordered couple of two vectors. Bilinear means separately linear on both $a$ and $b$. Symmetric means
$$\forall a,b:\ a\cdot b = b \cdot a.$$
Non-degenerate means
$$(\forall a:\ a \cdot b = 0)\ \Rightarrow\ b = 0.$$
Note: Alternative definitions are in use, somewhat different in some properties. Also the notation for inner product is not firmly established and may especially differ between mathematics and physics.
In a vector space with scalar product a covector can be defined using scalar product, as follows. If in $a\cdot b$ we keep $a$ fixed a linear function $\cV\to K$ is defined:
$$x: \cV \to K,\quad a \in \cV\ \mapsto\ 
x \cdot a \in K.$$
This member of $\cV^*$ we may denote by $a^*$. Here is the cause of the confusion, frequently done, between scalar product and dual space. 
The question arises: with the above definition of $a^*$ every element of $\cV^*$ is obtained varying $a$ in $\cV\,$? The answer is negative without further hypotheses on $\cV$ and on the scalar product.
But here I must stop.
A: There seems to be a confusion in your question. You speak of "scalar product" in a case that could potentially be non-symmetric (according to your description), but by definition, a scalar product is a bilinear symmetric positive definite form. So you must have misunderstood something in your original source of information. Do you have a reference that would help us to clarify the bit you actually don't understand. As it is, the question doesn't make much sense.
A: It's not a scalar product. By definition a 1-form $\omega$ takes a vector $v$ as input and gives a scalar $\omega(v)$ as output. As 1-forms and vectors are different entities, symmetry of the operation does not even make sense.
It is true however that the more symmetric notation $\langle\omega,v\rangle$ is sometimes used. It stems from the fact that if your manifold is equipped with a non-degenerate bilinear form $g$, typically a metric, you get an isomorphism between the space of 1-forms and that of vectors. To the form $\omega$ it associates the vector $\omega^\sharp$ defined $\omega(v)=g(\omega^\sharp, v)$ for any vector $v$. In components the isomorphism is just the "raising indices" operation you may be familiar with. If $g$ is a metric then it is symmetric and $g(\omega^\sharp, v) = g(v,\omega^\sharp)$. Writing $\langle, \rangle$ for $g$ and identifying forms and vectors because of the isomorphism leads to the notation $\langle\omega, v\rangle $.
I should note that the notation $\langle\omega,v\rangle$ is sometimes used even if no preferred isomorphism between vectors and 1-forms has been chosen. We may discuss wether or not it is good notation, in any case it does not represent a scalar product.
