I find it helpful to work through the definition of multiplication on $\mathcal{D}^{(m)}$ when $m = \infty$, in which case it reduces to the "classical" ring of differential operators in the sense of Grothendieck; read sections 16.7 and 16.8 of [EGA 4, Quatrième partie][1].

So let $A$ be a commutative base ring, let $S = Spec(A)$ and let $X = \mathbb{A}^1_S$ so that $B = A[t] = \Gamma(X, \mathcal{O})$. We want to work out $\mathcal{D}^{(\infty)}(X)$. Let $Y = X \times_S X$ and $m = \infty$.

Let $n \geq 0$. First we have to work out the global sections of the sheaf $\mathcal{P}^n_{Xm}(Y)$, considered as a $B$-module. This will turn out to be _dual_ (as a $B$-module) to the $B$-module of all Grothendieck differential operators of order at most $n$.

Now $\mathcal{O}(Y) = B \otimes_A B$ is isomorphic as an $A$-algebra to the polynomial ring $A[t,t']$ where $t \mapsto t \otimes 1$ and $t' \mapsto 1 \otimes t$. The diagonal immersion $X \hookrightarrow Y$ corresponds to the algebra surjection $B \otimes_A B \to B$ which is just the multiplication map. So the _ideal_ of the diagonal, namely the kernel of this map, is generated as an ideal by the element 

$$\tau := t \otimes 1 - 1 \otimes t.$$

Let's view $\mathcal{O}(Y)$ as a $B$-algebra via the map $b \mapsto b \otimes 1$; then $\mathcal{O}(Y) \cong B[\tau]$. By definition, the global sections of $\mathcal{P}^n_{X\infty}(Y)$ are just

$$P^n := \mathcal{O}(Y) / (\tau^n)$$

so in particular it is a free $B$-module of rank $n$ with generators (the images of) $\tau^i$ for $0 \leq i < n$. By definition, 

$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(\mathcal{O}(Y) / (\tau^n), B) =: D_n $$

which is again a free $B$-module of rank $n$; let $\{ \partial^{[i]}, i=0, \ldots, n-1\}$ be the dual basis for this module. 

Now the multiplication map $D_r \times D_s \to D_{r+s}$ is the $B$-module dual of the map $\delta : P_{r+s} \to P_s \otimes P_s$. This turns out to be a $B$-algebra homomorphism and its key property is that

$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$

(it is a "primitive element" in an appropriate bialgebra). Let's now work out how to multiply $\delta^{[i]}$ by $\delta^{[j]}$ (drop the bars for clarity):

$$(\delta^{[i]} \cdot \delta^{[j]})(\tau^k) =  (\delta^{[i]} \otimes \delta^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{l=0}^k \binom{k}{l} \delta^{[i]}(\tau^l) \delta^{[j]}(\tau^{k-l})$$

which is just $\binom{i+j}{i}\delta_{k,i+j}$. Since $\binom{i+j}{i} \delta^{[i+j]}$ has the same effect on each $\tau^k$, we deduce that

$$ \delta^{[i]} \cdot \delta^{[j]} = \binom{i+j}{i} \delta^{[i+j]}$$

which is hopefully the familiar rule for multiplying divided powers (morally $\delta^{[i]} = \delta^i/i!$). 

The point of the Berthelot construction is that it is possible to _vary_ the divided-power structure on the diagonal, and thereby control just how many divided powers one gets in $\mathcal{D}^{(m)}$. For example, if $m = 0$ then you instead allow _all_ divided powers on the ideal of the diagonal (algebraically this means you consider the divided power algebra of the ideal $(\tau)$ in $B[\tau]$ to get $\oplus_{n=0}^\infty B \tau^{[n]}$), and when you take the $B$-dual, these divided powers in $\tau$ "remove" the divided powers in $\partial$ and you end up with $\mathcal{D}^{(0)}(X) = B[\partial]$, the ring of _crystalline differential operators_ (no divided powers).

I hope that you can now find [the paper you posted][2] a little easier to read. It does explain the details of what happens for a given "level" $m$ quite well, in my opinion.


  [1]: http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1967__32_/PMIHES_1967__32__5_0/PMIHES_1967__32__5_0.pdf
  [2]: http://arxiv.org/abs/0811.1168