I am having trouble understanding several things in this proof:
Let $poly_D(\mathbb{F})$ be the space of all polynomials in the ring $\mathbb{F}[x_1,...,x_n]$ with degree $\leq{D}$, and let $P\in{poly_D(\mathbb{F})}$ be a univariate polynomial. Let us define a line $l$ as a 1-dimensional affine subspace of $\mathbb{F}^n$. We parametrize $l$ by a map $\gamma:\mathbb{F}\rightarrow\mathbb{F}^n$, defined by $\gamma(t)=at+b$ for vectors $a,b\in{\mathbb{F}^n}$, $a\neq0$. We define $Q(t)=P(\gamma(t))=P(at+b)$. We see that $Q(t)$ is a polynomial in one variable of degree $\leq{D}$. Since $P$ vanishes at $D+1$ points of $l$, $Q$ vanishes at $D+1$ values of $t$. Therefore, $P$ is the zero polynomial.
Not too sure why $l$ is being represented by $at+b$ instead of just $at$. Also, does $at+b$ represent each point of the line $l$?
As the line is an affine subspace--not necessarily a linear subspace--then it does not necessarily have the origin as one of its points. That's why we need the $+b$ in there. That can certainly be confusing, as "linear function" is often used to refer to a function whose image is a line.
Given any $t\in\Bbb F,$ $at+b$ is one of the points on the line $l.$ Moreover, given a point $x$ on the line $l,$ there is a $t\in\Bbb F$ such that $x=at+b.$ Put another way, $at+b$ is meant to represent an arbitrary point on the line $l,$ but not all of them at once.