Linear equation can be represented by a linear form, and its solution space is the same thing as kernel of this form. The same is true for system of linear equations.
But this lecture notes suggest that solution space also can be set in contra-variant way. Represented in exact sequence (in 3-dismensions) it looks like:
Equations -> <x,y,z> -> Solutions -> 0
there each element of 'Equations' basis maps to formal representation of some equation in <x,y,z>. So solution space will be quotient rather than kernel.
I can not understand why this is true. Why? Is such representation functional?
Let $V$ be a finite dimensional space and let $W \subseteq V$ be a subspace. Denote by $W^{0}$ the annihilator of $W$. Then we have a natural isomorphism $W^{*} \cong V^{*}/W^{0}$ obtained by mapping $[\varphi] \in V^{*}/W^{0}$ to the linear functional $\varphi|_{W}$.
Suppose we are given $k$ linear functionals $(\varphi)_{i=1}^k$ on $V$. We can define a map $T \colon V \rightarrow \mathbb{F}^k$ by $T(v) = (\varphi_1(v), \dots, \varphi_k(v))$ and then $\ker(T)$ describes the solution space of the system of linear equations
$$ \varphi_1(v) = \dots = \varphi_k(v) = 0. $$
Now, we can look at this from the dual point of view. Consider the dual map $T^{*} \colon \left( \mathbb{F}^k \right)^{*} \rightarrow V^{*}$. The natural basis on $\left( \mathbb{F}^k \right)^{*}$ is $e^i$ (the dual to the standard basis on $\mathbb{F}^k$) and we have $T^{*}(e^i) = \varphi_i,$ so the image of $T^{*}$ is precisely the subspace of equations that $\ker(T)$ satisfies. In other words, $\operatorname{Im}(T^{*}) = \ker(T)^0$ and so
$$ \operatorname{coker}(T^{*}) = V^{*}/\operatorname{Im}(T^{*}) = V^{*}/\operatorname{\ker(T)}^{0} \cong \ker(T)^{*}. $$
Thus, the dual to the solution set is naturally isomorphic to the cokernel of the dual map. If $V = \mathbb{F}^n$ is a free vector space, we can use the natural basis to identify $V$ with $V^{*}$ and also obtain $\ker(T) \approx \operatorname{coker}(T^{*})$ which is what is described in the document.
Let me do explicitly his example. We have $V = \mathbb{F}^3$ and $k = 3$ with
$$ \varphi_1(x,y,z) = x + y + z, \,\,\, \varphi_2(x,y,z) = 2x + y, \,\,\, \varphi_3(x,y,z) = 3x + 2y + z. $$
The map $T$ is given by $T(x,y,z) = (x + y + z, 2x + y, 3x + 2y + z)$ and $ \ker(T) = \left< (1,-2,1) \right>$. Here, $x,y,z$ can be thought of as coordinates on $V^{*}$ corresponding to the dual of the standard basis.
Dually, we can consider $V^{*}/ \left< \varphi_1, \varphi_2, \varphi_3 \right>$. If we identify $V$ with $V^{*}$, we see that this space is isomorphic to
$$ \mathbb{F}^3 / \left< (1,1,1), (2,1,0), (3,2,1) \right> = \left< [(1,-2,1) ]\right>. $$