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 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 (EGA IV, 16.7.1.1), the global sections of $\mathcal{P}^n_{X\infty}(Y)$ are just

$$P^n := \mathcal{O}(Y) / (\tau^{n+1})$$

--- this is the algebra of functions on the $n$-th infinitesimal neighbourhood of the diagonal $\tau = 0$ inside $\mathcal{O}(Y)$ (hence the $n+1$ in the exponent). So in particular it is a free $B$-module of rank $n+1$ with generators (the images of) $\tau^i$ for $0 \leq i \leq n$. By definition,

$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(P_n, B) =: D_n $$

which is again a free $B$-module of rank $n+1$; let $\{ \partial^{[i]}, i=0, \ldots, n\}$ 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 a map $\delta : P^{r+s} \to P^r \otimes P^s$ which is constructed in EGA IV, Lemma 16.8.9.1. Morally $\delta$ sends $a \otimes b$ to $a \otimes 1 \otimes 1 \otimes b$, as Gros/Le Stum/Quirros mention. This turns out to be a $B$-algebra homomorphism, and tts key property is that

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

(it is a "primitive element" in an appropriate bialgebra --- see EGA IV.4, 16.8.9.4). Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):

$$(\partial^{[i]} \cdot \partial^{[j]})(\tau^k) = (\partial^{[i]} \otimes \partial^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{a + b = k} \binom{k}{a} \partial^{[i]}(\tau^a) \partial^{[j]}(\tau^b)$$

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

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

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

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,

$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^n B \tau^{ \{ a \} } $$

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.

**Edit:** To see what the map $X \to P_{Xm}(Y)$ looks like in the case $X = \mathbb{A}^1_S$, it is enough to describe the corresponding map $C := \mathcal{O}(P_{Xm}(Y)) \to B = \mathcal{O}(X)$ on functions, because everything in sight is affine. $C$ is a $B$-algebra, generated by symbols $\tau^{ \{a \} }$ for all $a \geq 1$ subject to the relations

$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \frac{q_{a+b}!}{q_a!q_b!} \tau^{ \{ a + b \} }$$

(note that the structure constant $\frac{q_{a+b}!}{q_a!q_b!}$ is actually an *integer*, this again follows from Lemma 1.1.3(i) in Berthelot's paper.) The map $C \to B$ sends all of the generators $\tau^{ \{a \} }$ to zero, and the map $C \to P^n$ which corresponds to the closed subscheme $P^n_{Xm}(Y)$ of $P_{Xm}(Y)$ sends all of the $\tau^{ \{ a \} }$ to zero for $a \geq n+1$.

This description makes it easy to see that the algebra of functions $C$ on $P_{Xm}(Y)$ is isomorphic to the polynomial ring $B[\tau]$ when $m = \infty$ (since we can take $q_a$ to be always zero in this case), and to the "free" divided-power algebra $B[\tau^{[n]} : n\geq 1]$ that Gros/Le Stum/Quirros call $\Gamma_\bullet(B \tau)$ when $m = 0$. This is because $q_a = a$ in this case, so the defining relations between the $\tau^{ \{a \}}$ reduce to

$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \binom{a+b}{a} \tau^{ \{ a + b \} }.$$