I find it helpful to first 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.
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 $< n$.
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 $\delta : P^{r+s} \to P^r \otimes P^s$. 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):
$$(\delta^{[i]} \cdot \delta^{[j]})(\tau^k) = (\delta^{[i]} \otimes \delta^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{a + b = k} \binom{k}{a} \delta^{[i]}(\tau^a) \delta^{[j]}(\tau^b)$$
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).
Now to answer your question, let the level $m \geq 0$ be fixed. Then as Gros/Le Stum/Quirros explain just before Definition 2.5,
$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^{n-1} B \tau^{ \{ a \} } $$
where $\tau^{ \{ a \} }$ is a symbol that "behaves like $\tau^a / q_a!$" (where $q_a$ is the integer part of $a / p^m$: thus $a = q_a p^m + r_a$ say).
To understand the multiplication of the dual vectors to these $\tau^{ \{ a \} }$, namely the $\partial^{ \langle a \rangle }$, we need to understand how to comultiply the $\tau^{ \{ a \} }$. So we compute (again dropping bars for convenience)
$$ \delta( \tau^{ \{ a \} }) = \frac{1}{q_a!} \delta(\tau)^a = \sum_{i+j = a} \frac{1}{q_a!} \binom{i+j}{i} \tau^i \otimes \tau^j = \sum_{i+j = a} \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \tau^{ \{ i \} } \otimes \tau^{ \{ j \} }$$
and the same computation as above in the case $m=\infty$ shows that
$$ \partial^{\langle i \rangle} \cdot \partial^{\langle j \rangle} = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \partial^{\langle i + j \rangle}.$$
The interesting thing is that this structure constant $\frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i}$ is always a $p$-integral rational number (see Lemma 1.1.3(i) in Berthelot's paper), so it makes sense whenever $A$ is an algebra over $\mathbb{Z}_{(p)}$, say, and in particular if $A$ had characteristic $p$. Note that if $A$ was a $\mathbb{Q}$-algebra, then there would be a ring homomorphism from $\mathcal{D}^{(m)}$ to $A[t; \partial]$ which sends
$$ \partial^{\langle i \rangle} \mapsto \frac{q_i!\partial^i}{i!} $$
since
$$ \left(\frac{q_i! \partial^i}{i!}\right) \cdot \left(\frac{q_j! \partial^j}{j!}\right) = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \left( \frac{q_{i+j}! \partial^{i+j}}{(i+j)!}\right).$$
Thus morally $\partial^{\langle i \rangle}$ should be thought of as "modified divided powers" $q_i! \partial^i / i!$.
Finally, you don't need all of the $\partial^{\langle i \rangle}$ to generate $\mathcal{D}^{(m)}$. As is well-known, the full ring of Grothendieck differential operators in characteristic $p$ can be generated by the divided powers $\partial^{[p^a]}$ (for all $a \geq 0$). Since $q_i = 0$ for $i < p^m$ and $q_{p^m} = 1$, the modified divided powers $\partial^{\langle p^i \rangle}$ are equal to the "true" divided powers $\partial^{[p^i]}$ for $0 \leq i \leq m$. If $a > m$ then since $q_{p^a} = p^{a-m}$,
$$\partial^{\langle p^a \rangle} = \frac{ p^{a-m}! }{ p^a! } \partial^{p^a} = \left(\frac{ p^{a-m}! (p^m!)^{p^{a-m}} }{p^a!} \right) (\partial^{\langle p^m \rangle})^{p^{a-m}}$$
shows that $\partial^{\langle p^a \rangle}$ is a $p$-adic unit times a power of $\partial^{\langle p^m \rangle}$ for $a \geq m$ since the $p$-adic valuation of that big fraction is
$$\frac{p^{a-m}-1}{p-1} + p^{a-m} \frac{p^m-1}{p-1} - \frac{p^a-1}{p-1} = 0.$$
So we see that $\mathcal{D}^{(m)}(\mathbb{A}^1_S)$ in this case is the $A$-algebra generated by $t$ and the divided powers $\partial^{[p^0]}, \partial^{[p^1]}, \ldots, \partial^{[p^m]}$, subject to appropriate natural relations.