on the definition of graded Betti numbers Let's use as reference the slides 19-31.
Let $S=k[x_1,\cdots,x_n]$ and $M$ a finitely generated graded $S$-module. Then by Hilbert's Syzygy Theorem, $M$ has a minimal, graded, free resolution of length at most $n+1$, i.e.
$0 \rightarrow F_m \rightarrow F_{m-1} \rightarrow \cdots \rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$, where $F_i$ is free, graded $S$-module and $m \le n+1$.
Towards introducing the graded Betti numbers, let $F_i = \bigoplus_{d} (F_i)_d$, where $(F_i)_d$ is the homogeneous component of degree $d$. Now, $(F_i)_d$ is a finite-dimensional $k$-vector space. 
Here is the part that confuses me: at slide 20, the authors says "we give to each generator of $F_i$ a degree" and subsequently writes $F_i = \bigoplus_{j \in \mathbb{Z}} S(-j)^{\beta_{i,j}}$, and he mentions that $S(-j)$ is cyclic free module. 
Question 1: I can not understand what it means to give a degree to each generator (i don't see why every generator has to be inside some homogenous component, in which case we could associate a degree to each generator).
Question 2: What exactly is $S(-j)$? Initially, i thought that $S(-j)$ was the homogeneous component of $S$ of degree $-j$, however, this is not a cyclic module as the author claims at slide 27.
Any examples will be highly appreciated.
 A: First, Question 2: For any graded object $M=\bigoplus M_d$ and integer $j$, one gets a new graded object $M(-j)$ with grading obtained by shifting that of $M$: the degree $d$ piece of $M(-j)$ is by definition $M(-j)_d=M_{d-j}$. 
Now for Question 1, since the module $F_i$ is free it can be written as a direct sum 
$$F_i=S^{\oplus \beta_i}$$ of some number of copies of the free module of rank one $S$. But to make the maps in your resolution graded of degree zero (a map $\phi:M \rightarrow N$ is graded of degree $0$ if $\phi(M_d)\subseteq N_d$), you may have to shift the gradings on the summands $S$ that appear. Equivalently, you may have to give the generator $1 \in S$ a non-zero degree, in order to map it onto something with non-zero degree---but this is just a book-keeping device, as I hope will become clear.
I'll give two examples now, which are the first two everyone learning this stuff for the first time should work out:
First, assume $n=1$ and $M=k=k[x]/(x)$ is the trivial module concentrated in degree $0$. There is a free resolution
$$0 \rightarrow S \rightarrow S \rightarrow M \rightarrow 0$$ of length $0$, where the map from $S$ to itself is multiplication by $x$. In order to make this a degree $0$ map of graded modules, one must shift the grading, replacing the left-hand $S$ by $S(-1)$ (since the degree $1$ piece of $S(-1)$ is the constants $k$, which get mapped onto constant multiples of the degree $1$ element $x \in S$, this makes good sense). So this resolution is usually written
$$0 \rightarrow S(-1) \rightarrow S \rightarrow k \rightarrow 0.$$ 
The second example to check out is again $M=k$, but this time with the number of variables $n=2$. This time there is a resolution of length $2$,
$$0 \rightarrow S(-2) \rightarrow S(-1)^{\oplus 2} \rightarrow S \rightarrow k \rightarrow 0,$$ where the map $S(-1)^{\oplus 2} \rightarrow S$ is given by $(f,g) \mapsto xf+yg$ and the map $S(-2) \rightarrow S(-1)^{\oplus 2}$ is $f \mapsto (yf,-xf)$.
This fits into a family of examples (one for each $n=1,2,3,...$) called the Koszul resolutions. These are in some sense the simplest nontrivial free resolutions.
