I am not clear on how why this proof works. I understand that for any $p$ this algorithm gives us back a $\overline{p}$ such that: $p$ and $\overline{p}$ are congruent mod $I$, and only standard monomials appear in $\overline{p}$.
My question is why do the standard monomials span $C[x]/I$, and why are the linearly independant?
Also in the last comment "zero cannot me written as a nontrival combination of standard monomials in $C[x]/I$", I know that standard monomials in $C[x]$ are the monomials that are not in $init(I)$, by standard monomials in $C[x]/I$ do they mean the monomial we get after applying this algorithm?
Let $f,g \in \mathbb{C}[x_1,\ldots,x_n]$ be two polynomials.
By definition of the quotient ring, $[f] = [g]$ in $\mathbb{C}[x_1,\ldots,x_n]/I$ if and only if $f-g \in I$.
Fix a monomial ordering on $\mathbb{C}[x_1,\ldots,x_n]$ and a Gröbner basis of $I$. Since for any $p \in \mathbb{C}[x_1,\ldots,x_n]$ we have $[p] = [\bar{p}]$ where either $\bar{p}=0$ or only standard monomials appear in $\bar{p}$, it follows that $\mathbb{C}[x_1,\ldots,x_n]/I$ is spanned by (equivalence classes of) standard monomials.
Suppose $[f]=[g]$ and $[f],[g]$ are different linear combinations of (equivalence classes of) standard monomials. Then $[0] = [f]-[g]$ is a nontrivial linear combination of (equivalence classes of) standard monomials. That would mean a nontrivial linear combination of standard monomials belongs to $I$. But this is impossible, because every nonzero element of $I$ must contain a nonstandard monomial.
So every element of $\mathbb{C}[x_1,\ldots,x_n]/I$ is a unique linear combination of (equivalence classes of) standard monomials, and hence the (equivalence classes of) standard monomials form a basis of $\mathbb{C}[x_1,\ldots,x_n]/I$.
For your last question:
Here "standard monomials" refers to the usual notion; the clause "in $\mathbb{C}[x_1,\ldots,x_n]/I$" refers to "written". That is, when you take the statement
and reduce it modulo $I$, then you get