For any graded module $M$ we denote $M(a)$ the module $M$ "shifted by $a$" so that $M(a)_d=M_{a+d}$. Thus for example the free $S$-module of rank $1$ generated by an element of degree $a$ is $S(-a)$.
Here $S=k[x_1,x_2,\dots,x_n]$, where $k$ is a field.
I don't understand the statement in bold font. We know that $S=M_0\oplus M_1\oplus\dots\oplus M_n$ where $M_d$ is the group of homogeneous polynomials of degree $d$. Hence $S(-a)=M_{-a}\oplus M_{-a+1}\oplus \dots\oplus M_{n-a}$
What does this even mean?
That's almost right. But, note that $$ S = \bigoplus_{d=0}^{\infty}M_d = M_0 \oplus M_1 \oplus \ldots $$
not the finite direct sum that you wrote.
But the next thing you wrote is a little more wrong. The point is that after shifting, the $d$ degree part of $S(-a)$ is whatever the $d-a$ graded part of $S$ is, ie. $M_{d-a}$ if $a\leq d$ or $0$ otherwise. But don't forget, at this point, we're talking about $S(-a)$ as a module over $S$, not as a ring. It IS in fact a ring, but we're thinking of it as the free module with generator in degree $a$ in this case.