See here. Does it follow immediately that $\mathcal{D}_A(M, M)$ as defined in the link is a filtered, almost commutative ring? How can I visualize this geometrically?
Not so immediately, but rather easily (as far as the modules are of a "good" kind, i.e., projective and finitely generated). It all boils down to prove that, if $$ D\in\mathcal{D}^n_A(B):=\mathcal{D}^n_A(B,B)\, , $$ then the map $$ B\times\cdots\times B\ni(b_1,\ldots,b_n)\stackrel{\mathrm{S}^n(D)}{\longrightarrow}[b_1\cdots [b_n,D]\cdots]\in\mathrm{End}_B(B)\equiv B $$ is $A$multilinear and, with respect to each entry, it fulfils the Leibniz rule. Since a derivation factors through a $B$linear map on the $B$module of differential oneforms $\Lambda^1B$ (in analogy with the "jet modules" I mentioned before), the "$n^\textrm{th}$ symbol" $\mathrm{S}^n(D)$ of $D$ may be also thought of as an element of the $B$module $$ \mathrm{Sym}^n(\Lambda^1B) \, . $$ Now, to check your claim (EDIT: which is certainly true for $M=B$, see Michael Bachtold's comments below), you need two obvious properties:
1) $D\in\mathcal{D}^n_A(B) $ if and only if $\mathrm{S}^{n+1}(D)=0$;
2) if $D_1\in\mathcal{D}^{n_1}_A(B) $ and $D_2\in\mathcal{D}^{n_2}_A(B) $, then $\mathrm{S}^{n_1+n_2}(D_1\circ D_2)=\mathrm{S}^{n_1}(D_1)\cdot \mathrm{S}^{n_2}(D_2)$, where "$\cdot$" is the product in the graded commutative algebra $\mathrm{Sym}(\Lambda^1B)$.
The general case, i.e., when $D\in \mathcal{D}^n_A(M):=\mathcal{D}^n_A(M,M)$, can be reduced to the previous one by using the identifications $$ \mathcal{D}^n_A(M)\ni D\longleftrightarrow \widehat{D}\in\mathrm{Hom}_B(M, \mathcal{D}^n_A(B,M))\, , $$ where $\widehat{D}(m)(b):=D(bm)$, and $\mathcal{D}^n_A(B,M)\equiv \mathcal{D}^n_A(B)\otimes_B M$, which imply $\mathcal{D}^n_A(M)\equiv \mathcal{D}^n_A(B)\otimes_B\mathrm{End}_B(M)$, but commutativity is not guaranteed (again, see comments below).
If the modules are of a "bad" kind, then I wouldn't know how to deal with them!

2$\begingroup$ Wouldn't almost commutative require that any two zero order operators $D_1,D_2:M\to M$ commute? $\endgroup$ – Michael Bächtold Jul 7 '15 at 8:14

1$\begingroup$ Here's an explicit counterexample: $B=C^\infty(\mathbb{R})$, $M=B\oplus B$ and consider the first order operators $D_1: (f,g)\mapsto (\partial f,g)$ and $D_2: (f,g)\mapsto (\partial g, \partial f)$. Now $[D_1,D_2]$ is of order 2 and not 1, hence $\mathcal{D}(M,M)$ is not almost commutative. So finitely generated an projective is not enough. Maybe invertible $M$ will do. $\endgroup$ – Michael Bächtold Jul 7 '15 at 9:44
In order for $\mathcal{D}_A(M,M)$ to be almost commutative we need in particular that $\mathcal{D}_A^0(M,M)$ be commutative. But this is also sufficient since the proof of commutativity of the associated graded proceeds by a simple induction on $\mathrm{ord}\, D_1 + \mathrm{ord}\, D_2$ using the Jacobi identity in the form $[b,[D_1,D_2]]=[[b,D_1],D_2]+[D_1,[b,D_2]]$. So finitely generated projective $M$ is not enough as suggested by the previous answer.
If you are interested in the question of characterising those modules $M$ with commutative endomorphism rings here is something:
Concerning the geometric interpretation of the graded in the case $M=B$ (which is also supplied with a Poisson bracket), this is the wellknow cotangent space of $\mathrm{Spec} B$ as you might see from G_infinity's answer. You can find out more about it in the literature on Dmodules, microlocal analysis or quantization.