I am following a presentation, which says that for an affine set $L \subseteq \mathbb{R}^n$ it is: $$L=\left\{x|Ax=b \right\}$$
for some $A,b$.
The first definition of $L$ as an affine set is given as; given two points $x$ and $y$ with $x,y \in L$, it is is always $\alpha x + \beta y \in L$ where $\alpha,\beta \in \mathbb{R}$ and $\alpha + \beta = 1$. This is very clear and straightforward. But I did not get its relation with $L=\left\{x|Ax=b \right\}$. As far as I remember from my linear algebra classes which I took years ago, $Ax=b$ is a non-homogeneous system of linear equations. But I cannot see how it is connected with the definition of an affine set. I need some insight here.
You have two definitions of affine, one that is given via generators:
$L$ is affine if there are $A,c$ with ${x:Ax=c}$ (the set of all affine sets is generated by all $A,c$).
One that is given via constraints:
$L$ is affine if for every $a,b\in\mathbb{R}$ with $a+b=1$ and $x,y\in L$ there is also $ax+by\in L$ (this is a constraint on a subset of $\mathbb{R}^n$).
You need to show that every set satisfying one definition also satisfies the other. The easier direction is showing that something generated via $A,c$ has the latter property. The harder direction is finding generators $A,c$ for a set $L$ satisfying the constraint. By the way, I personally favor another definition:
$L$ is affine if there is a linear subset $K$ and a vector $c$ such that $L=\{x+c|x\in K\}$. You should look up your linear algebra results for describing a linear set $K$ via either its generators or constraints, and for converting between the two descriptions. Can you show that for any $c\in L$ the set $K=\{x-c|x\in L\}$ is linear for either definition of affine linear?
Specifically, in terms of your question, a set $K=\{x|Ax=0\}$ and a set $K'$ such that for all $a,b\in\mathbb{R}$ and all $x,y\in K'$ the sum $ax+by$ is again in $K'$ are both linear.