Why do we care about the annihilator of a row space? It's natural to find the subspace annihilated by the set of linear functionals because the subspace in this case conceptually represents the solution space. However, what does it mean to find the annihilator of a row space? I think the row space is the one that should be treated as the annihilator—because $(A_{i1},...,A_{in})$ is uniquely paired with linear functionals $f_{i}(x_{1},...,x_{n})=A_{i1}x_{1}+\cdots +A_{in}x_{n}.$ If we want to find $\alpha$ such that $$f_{i}(\alpha)=0\quad i=1,...,m$$ we need to focus on finding the subspace annihilated by a row space, not the annihilator of a row space, I suppose. Here's the quote from the book that puzzled me a lot:

Now one may look at the system of linear equations from the 'dual' point of view. That is, suppose that we are given $m$ vectors in $F^{n}$ $$\alpha_{i}=(A_{i1},\ldots,A_{in})$$ and we wish to find the annihilator of the subspace spanned by these vectors. Since a typical linear functional on $F^{n}$ has the form $$f(x_{1},\ldots,x_{n})=c_{1}x_{1}+\cdots+c_{n}x_{n}$$ the condition that $f$ be in this annihilator is that $$\sum_{j=1}^{n}A_{ij}c_{j},\quad i=1,\ldots,m$$ that is, that $(c_{1},\ldots,c_{n})$ be a solution of the system $AX=0.$ From this point of view, row-reduction gives us a systemic method of finding the annihilator of the subspace spanned by a given finite set of vectors in $F_{n}.$

 A: Under the common conventions, we think of column vectors as elements of the vector space $\mathbb{F}^n$ and of row vectors as elements of the dual space $(\mathbb{F}^n)^{*}$. Each row vector $v = (x_1, \dots, x_n)$ defines uniquely a linear functional $\varphi_v \colon \mathbb{F}^n \rightarrow \mathbb{F}$ on $\mathbb{F}^n$ by
$$ \varphi_v \left( \begin{pmatrix} y_1 \\ \dots \\ y_n \end{pmatrix} \right) = (x_1, \dots, x_n) \begin{pmatrix} y_1 \\ \dots \\ y_n \end{pmatrix} = \sum_{i=1}^n x_i y_i $$
and all linear functionals are obtained in this way.
A subspace $V \subseteq \mathbb{F}^n$ can be described in two ways - as a span of a set of vectors ("a parametric description") or as a solution set of a linear system of equations ("an implicit description"). Passing from one description to the other involves calculating a basis of an annihilator of the appropriate set of row / column vectors.
Let me demonstrate this with an example. Consider the equation $x + y + z = 0$ on $\mathbb{F}^3$. The equation defines a two-dimensional subspace $V$ of $\mathbb{F}^3$ which is the same as the kernel of the linear functional $\varphi_{(1,1,1)}$ or the annihilator of the set $\{ (1,1,1) \}$ (where we identify $\left( \left( F^{n} \right)^{*} \right)^{*}$ with $\mathbb{F}^n$ so the annihilator is a subspace of column vectors). In this case, a basis of the annihilator is given for example by
$$ \operatorname{Ann}( (1,1,1) ) = \operatorname{span} \left \{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right \}. $$
This gives us a parametric description of $V$ as a span of two linearly independent vectors:
$$ V = \left \{ t \cdot \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \, : \, t,s \in \mathbb{F} \right \}. $$
Going the other way around, given a parametric description of $V$, we can calculate a basis for the annihilator of the spanning set of $V$ and obtain an implicit description of $V$ as a solution set of linear equations:
$$ \operatorname{Ann} \left( \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right) = \operatorname{span} \{ (1,1,1) \}. $$
To summarize, calculating a basis for the annihilator of a set of row vectors is the same as finding a basis for the solution set defined by a system of equations (linear functionals) corresponding to the row vectors. Calculating a basis for the annihilator of a set of column vectors is the same as a finding a basis for the set of equations (linear functionals) the vectors satisfy resulting in a description of the span as a solution set of a linear system of equations. 
