Let $B$ be a commutative $A$-algebra, and let $M$, $N$ be two $B$-modules. We can talk about the set of $A$-linear module homomorphisms $M \to N$, i.e. the set $\text{Hom}_A(M, N)$. Differential operators of order zero should be the $B$-linear maps from $M$ to $N$, i.e. $\text{Hom}_B(M, N)$.

First, note that the commutator $[f, b]$ (where $b \in B$) is a well-defined morphism $M \to N$. Then we make our first definition, the "Weyl Algebra" one.

Definition 1 (Weyl Algebra).Let $\mathcal{D}_A^0(M, N) = \text{Hom}_B(M, N)$. Define $$\mathcal{D}_A^n(M, N) = \{f \in \text{Hom}_A(M, N) \text{ such that }[f, b] \in \mathcal{D}_A^{n-1}(M,N)\}.$$We set $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0}\mathcal{D}_A^n(M, N)$.

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$. We now have our "Crystalline" definition.

Definition 2 (Crystalline).Let $I$ be the kernel of the diagonal map (i.e., the map $B \otimes_A B \to B, \ b \otimes b' \mapsto bb'$). Then $D: M \to N$ is said to be a differential operator of order $\le n$ if $\overline{D}$ annihilates $I^{n+1} \otimes_B M$. Let $\mathcal{D}_A^n(M, N)$ be the $B$-module of differential operators of order $\le n$. We define $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0} \mathcal{D}_A^n(M, N)$.

My question is, what is the easiest way to see that/the intuition behind the definitions of rings of differential operators between modules given above are equivalent?

**EDIT:** In the comments, Michael Bächtold is asking me to spell out the definition of $\delta_{B/A}$ and $\overline{D}$.

So say we have $B$ a commutative $A$-algebra. We want to formalize the notion of an $A$-linear endomorphism of $B$ which is ``close" to being $B$-linear. Let $D: B \to B$ be an $A$-linear endomorphism of $B$. Using $D$, we obtain a map$$\tilde{D}: B \otimes_A B \to B$$defined by $\tilde{D}: b \otimes b' \mapsto bD(b')$, which can also be viewed as a map$$\overline{D}: B \otimes_A B \otimes_B B \to B,$$where we have identified $B$ and $B \otimes_B B$ and the map is defined by $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$. Let us define $\delta_{B/A} = B \otimes_A B$. Then, we have a map:$$\overline{D}: \delta_{B/A} \otimes_B B \to B.$$

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$ defined by the same formula as above (that is, $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$ for $b \in B$, $b' \in B$ and $b'' \in M$). We now have our "Crystalline" definition.

In the quoted text, the inducing is in perfect analogy to what I wrote above.