This is right out of Kassel's Quantum Groups book which I am self-studying. It is on page 14.

The general set-up is this. Let $$A$$ be a filtered algebra with filtration $$F_0(A) \subset F_1(A) \subset \cdots \subset A$$, and suppose that $$I$$ is a two-sided ideal of $$A$$. The quotient algebra is then filtered with filtration $$F_i(A/I) = F_i(A)/F_i(A) \cap I$$. We know that for a filtered algebra, $$A$$, there exists an associated graded algebra, called $$\mathrm{gr}(A) = \oplus S_i$$ where $$S_i = F_i(A)/F_{i-1}(A)$$.

Define $$M(2)$$ as the polynomial algebra $$k[a,b,c,d]$$ and define $$SL(2) = M(2)/(ad - bc -1)$$.

His first claim is that $$\mathrm{gr}(A/I) = \oplus_{i \in \mathbb{N}} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I)$$. Following this, he claims that $$\mathrm{gr}(SL(2)) \cong k[a,b,c,d]/(ad-bc)$$ (note that the ideal $$(ad-bc-1)$$ is not generated by homogeneous elements so that $$SL(2)$$ is not graded).

I am unsure how one gets to these results, and any feedback on this would be greatly appreciated!

We have \begin{align*} \text{gr}(A/I) &=& \bigoplus F_i(A/I)/F_{i-1}(A/I) \\ &\cong& \bigoplus (F_i(A)/F_i(A) \cap I)/(F_{i-1}(A)/F_{i-1}(A) \cap I) \\ &\cong& \bigoplus F_i(A)/(F_{i-1}(A) + F_i(A) \cap I) \end{align*}
which follows from the fact that $F_{i-1}(A) \cap I \subseteq F_i(A) \cap I$ and I suppose the third isomorphism theorem.
With $A = k[a, b, c, d]$ and $I = (ad - bc - 1)$ we have that $F_i(A)$ consists of polynomials of total degree at most $i$ and that $F_{i-1}(A) + F_i(A) \cap I$ consists of polynomials of total degree at most $i-1$ together with polynomials of the form $f (ad - bc - 1)$ where $f$ has total degree at most $i-2$. Since $f \in F_{i-1}(A)$, instead of adding in terms of the form $f(ad - bc - 1)$ we can add in terms of the form $f(ad - bc)$. Do you see how the result follows from here?