Meaning of $\mathrm{d}_{B/A} \otimes 1$ in EGA IV and pullback of connections.

Question

I am reading some parts of EGA IV and come to the following question.$\require{AMScd}$

Suppose we have a commutative diagram $$\begin{CD} X' @>>> X\\ @VVV @VVV\\ S' @>>> S \end{CD}$$ of schemes. In EGA IV$_4$, (16.4.3.7), there is a map $$ \mathrm{d}_{X/S} \otimes 1: \mathcal{O}_{X'} \longrightarrow \Omega_{X/S}^1 \otimes_{\otimes_X} \mathcal{O}_{X'}.$$ And the same map in local case $$ \mathrm{d}_{B/A} \otimes 1: B' \longrightarrow \Omega_{B/A}^1 \otimes_B B'$$ also appears in EGA $0_\text{IV}$, (20.5.4.2), given a commutative diagram of rings $$\begin{CD} B' @<<< B\\ @AAA @AAA \\ A' @<<< A. \end{CD}$$

I am wondering how was this map defined. For example, let's consider the local case. Here are some of my thoughts.

1. The map $\mathrm{d}_{B/A} \otimes 1$ should be $A'$-linear as stated before the diagram EGA $0_\text{IV}$, (20.5.4.2).
2. If we consider $B \otimes_B B' \cong B'$ with $b \otimes b' \mapsto bb'$, and map $b \otimes b' $ to $\mathrm{d}_{B/A} (b) \otimes b'$, which is literally $\mathrm{d}_{B/A} \otimes 1$, but this will result in the zero map as we can always write $b \otimes b' = 1 \otimes bb'$ and $\mathrm{d}_{B/A}(1) = 0$. This is not what is expected, I think.
3. We know that $\mathrm{d}: B \to \Omega_{B/A}^1$ is an $A$-linear map. I didn't find a way to pull back this map to a $A'$-linear map $B' \to \Omega_{B/A}^1 \otimes_B B'$.
4. Acutually, the real problem I am considering is to pullback a connection functorially. That is, suppose we have a quasi-coherent sheaf $\mathcal{E}$ over $X$ together with a $S$-connection $$\nabla: \mathcal{E} \to \Omega_{X/S}^1 \otimes_{\mathcal{O}_X} \mathcal{E}.$$ I expect to get canonically a $S'$-connection $$f^\ast \nabla: f^\ast \mathcal{E} \to \Omega_{X'/S'}^1 \otimes_{\mathcal{O}_{X'}} \mathcal{O}_{X'}$$ on $f^\ast \mathcal{E}$, where $f$ is the map $X' \to X$. Now the pullback of $\mathrm{d}_{X/S}$ is the trivial case. But I didn't manage it using this perspective. However, if using that $\nabla$ is equivalent to an isomorphism $\epsilon: (p_1^{(1)})^\ast \mathcal{E} \to (p_2^{(1)})^\ast \mathcal{E}$ over the first infinitesimal neighborhood $X^{(1)}$ of $X$ in $X \times_S X$. Then the pullback is obvious. The map $(p_i^{(1)})$ are the projections $X^{(1)} \to X \times_S X \to X$, $i = 1, 2$ as introduced in EGA IV$_4$ (16.7.1). This pullback was also mentioned in Katz, Nilpotent connections and the monodromy theorem, (1.1.6). But I didn't understand his notation and he didn't explain the details.

Any comment is welcome. Thanks！