Let $\pi:E\to M$ be a smooth vector bundle. Then we have the following exact sequence $$ 0\to VE\xrightarrow{} TE\xrightarrow{\mathrm{d}\pi}\pi^*TM\to 0 $$ Where $VE$ is the vertical bundle. This sequence splits, so there is a bundle morphism $\delta:\pi^*TM\to TE$ such that $\mathrm{d}\pi\circ\delta=\mathrm{Id} \vert_{TE}$.
I have read that a choice of $\delta$ which splits this sequence is equivalent to a choice of a covariant derivative $\nabla$ on $E\to M$.
Could anyone please explain the correspondence between splitting the above sequence and between covariant derivatives on $E$?
The differential of a section $s: M \rightarrow E$ is the linear bundle map $$ s_*: T_*M \rightarrow T_*E. $$
A connection defines a directional derivative of a section $s$ at $p \in M$ in the direction $X \in T_pM$, denoted $$ \nabla_Xs(p) \in E_p \simeq V_{s(p)}E.$$ It is defined by $$\nabla_Xs(p) = P_{s(p)}(s_*X),$$ where $P: T_*E \rightarrow VE$ is a projection map. Thus, it is defined by a splitting of the sequence.