An Ehresmann connection on a vector bundle $\pi : E \to X$ is a splitting of the sequence,
$$(1) 0 \to V \to TE \to \pi^* TX \to 0 $$ which respects the linear structure on $E$ (meaning the section is invariant under the induced automorphism of $T E$ induced by scaling).
In algebraic geometry, I am familiar with the equivalence between connections on a locally free sheaf $\mathcal{E}$ and splitting of the sequence
$$(*) 0 \to \Omega_X^1 \otimes \mathcal{E} \to J^1(\mathcal{E}) \to \mathcal{E} \to 0 $$ I have heard it said that this is a version of the Ehresmann connection formalism but I am not able to make this precise. If I dualize the top sequence and use the fact that $V \cong \pi^* E$ then I recover,
$$ 0 \to \pi^* T^* X \to T^* E \to \pi^* E^* \to 0 $$ which looks similar to the jet bundle sequence. However, I am not sure how to directly compare these two sequences.
Furthermore, the notion of an Ehresmann connection makes perfect sense in the algebraic category. However, (at least without directly relating it to the jet bundle sequence) I do not see how to show that the data of a splitting recovers an (algebraic) connection.
The usual construction of a connection from an Ehresmann connection goes through algebraically. Call the splitting $v : T E \to \pi^* E$. Then given a section $s : X \to E$ we get $\mathrm{d}{s} : T X \to s^* T E$ and then $s^* v \circ \mathrm{d}{s}$ is a linear map $T X \to E$ defining $X \mapsto \nabla_X s$ thus defining the connection.
However, to reverse this process, it seems that I need to be able to choose, locally, flat sections for a connection $\nabla$ in order to define the kernel of $v$ which is the horizontal subspace. Does this mean the Ehresmann connection is really a transcendental object?
Question: "What exactly is the relationship between an Ehresmann connection and splitting of the jet sequence?"
Answer: Let $M$ be a finite rank projective $A$-module and assume $p: A^n \rightarrow M^*$ is a surjective map of left $A$-modules. We get a surjection $R:=A[x_1,..,x_n] \rightarrow Sym_A^*(M^*)$ and an isomorphism $R/I \cong B:=Sym_A^*(M^*)$. We get a sequence $A \rightarrow R \rightarrow B$ of rings and an exact sequence of $B$-modules
$$ I/I^2 \rightarrow B\otimes_R \Omega^1_{R/A} \rightarrow \Omega^1_{B/A} \rightarrow 0.$$
Let $E:=Spec(B)$ and let $X:=Spec(A)$. We get canonical maps
$$E \rightarrow^i \mathbb{A}^n_X:=\mathbb{A}^n \times X \rightarrow^q X$$
where $q$ is the projection map. The map $i$ is a closed immersion and it realize $E$ as a closed subscheme of the trivial vector bundle $\mathbb{A}^n_X$. If $k \subseteq A$ is a sub ring of $A$ with $S:=Spec(k)$ you get an exact sequence
$$(A) B\otimes_A \Omega^1_{A/k} \rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A} \rightarrow 0.$$
Note: This sequence is mentioned in Matsumuras book: If $B/A$ is "$0$-smooth" it follows $(A)$ is split exact.
Dualizing $(A)$ you get the exact sequence
$$ 0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow Hom_B(B\otimes_A \Omega^1_{A/k},B)$$
and when $A \rightarrow B$ is flat you get an isomorphism
$$ Hom_B(B\otimes_A \Omega^1_{A/k},B) \cong B \otimes_A Der_k(A).$$
giving an exact sequence (the "relative tangent sequence" of $B/A$)
$$ 0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow B\otimes_A Der_k(A).$$
The sheafification of this sequence gives the sequence
$$0 \rightarrow T_{E/X} \rightarrow T_{E/k} \rightarrow g^*(T_{X/k})$$
where $T_{E/X}$ is the relative tangent bundle and $g: E \rightarrow X$ is the projection map for $E$.
Answer: There is no immediate relation between this sequence and the Atiyah sequence. You should make the sequence in $(1)$ more explicit - what is $V$?
There is another sequence (see this link to an MO post)
https://mathoverflow.net/questions/385099/when-do-flat-holomorphic-connections-exist
which is the obstruction for the existence of a flat connection. When you have a connection $D$ on $M$ there is an extension (similar to the Atiyah extension)
$$(**) 0 \rightarrow End_A(M) \rightarrow L(M,D) \rightarrow^u Der_k(A) \rightarrow 0$$
which "splits" iff $M$ has a flat connection. Hence there are two fundamental sequences: The Atiyah sequence and the sequence $(**)$. By definition $L(M,D):=End_A(M) \oplus L$ with $L:=Der_k(A)$ and with the following Lie product:
$$[(f,x),(g,y)]:=([f,g]+[D(x),g]-[D(y),f]+R_D(x,y), [x,y])$$ where $R_D$ is the curvature of $D$. The map $u$ is the canonical map. You may check that $u$ is split iff there is a flat connection $D'$ on $M$. You get an extension class
$$c(M) \in Ext^1(L, End_A(M))$$
which is trivial iff $M$ has a flat connection.
Note moreover: The Atiyah sequence may be defined as follows: Let $J^1(M):= \Omega^1_{A/k}\otimes_A M \oplus M$ with the following left $A$-module structure:
$$ a( x\otimes e,f):= ((ax)\otimes e+d(a)\otimes f, af).$$
This gives a sequence which is split iff M has a connection
$$ D: M \rightarrow \Omega^1_{A/k} \otimes_A M.$$
If $L:=Der_k(A)$ there is dually an "Atiyah sequence" defined as follows: Let $J_1(M):=L\otimes_A M \oplus M$ with the "obvious" module structure. You get an exact sequence
$$ 0 \rightarrow M \rightarrow J_1(M) \rightarrow L\otimes_A M \rightarrow 0$$
which is split by a connection $D: L \rightarrow End_k(M)$.
If you look at the arXiv preprint server you find these types of structures filed under the name "Lie-algebroid" or "Lie-Rinehart algebra". There are many papers published on this subject.