I was taught, in different courses, the following definitions of connection.
An Ehresmann connection on a fibre bundle $\pi:B\rightarrow M$ is a distribution $H\hookrightarrow TB$ such that: $$\forall b\in B\quad T_bB=H_b\oplus V_b(\pi)$$ where $V_b(\pi)\subset TB$ is the subspace of vertical vectors at $b\in B$: $$V_bB=\{v\in T_bB| T_b\pi(v)=0\in T_{\pi(b)}(M)\}$$ We can identify an Ehresmann connection with a map $\omega: TM\rightarrow TB$ as follows Fix $H$, $u\in T_bB$, $v=T_b\pi(u)\in T_{\pi(b)}M$. Then $u\equiv u_{(H)}\oplus u_{(V)}$. Call $\tilde{v}:=u_{(H)}$ the horizontal lift of $v$, this is uniquely defined: $u'\in(T_b\pi)^{-1}(v)\Rightarrow T_b\pi(u-u')=v-v=0 \Rightarrow u-u'\in V_b(\pi)\Rightarrow u_{(H)}=u'_{(H)}$
Consequently an Ehresmann connection defines $\omega_H:TM\rightarrow TB$, fiberwise: $$\forall b\in B,\ (\omega_H)_b:T_{\pi(b)}M\rightarrow T_bB\quad\ v\in T_{\pi(b)}M\mapsto \tilde{v}\in T_bB$$ such that $T_b\pi\circ\omega_H=id_{TM}$.
Conversely,any such map determines an Ehresmann connection $H_\omega$, by setting $$(H_\omega)_b =\{u\in T_bB:\ u\in Im(\omega_b)\}$$ Such a map $\omega$ makes the following Witney sequence split: $$0\longrightarrow VB(\pi)\longrightarrow TB\xrightarrow{T\pi}TM\longrightarrow 0$$ Now given a (local) vector field $\xi\in\chi(U)$, we can lift its values through $\omega$ to a vector field $\xi_H$ valued in $TB$, in coordinates: $$\xi_H(x,y)=\xi^\mu(x)(\partial_{x^\mu}(x)-\Gamma^i_\mu(x,y)\partial_{y^i}(x,y))$$ Recalling that upon fibred coordinate change $(x,y)\mapsto(x'(x),y'(x,y))$: \begin{cases} \partial_{x^\mu}=J^\nu_\mu\partial_{x'^\nu}+J^i_\mu\partial_{y'^i}\\ \partial_{y^i}=J^j_i\partial_{y'^j} \end{cases} Thence an Ehresmann connection is determined by: \begin{cases} \Gamma^i_\mu:B\rightarrow\mathbb{R}\\ \Gamma'^i_\mu(x',y')=\bar{J}^\nu_\mu((J^i_j(x,y)\Gamma^j_\nu(x,y)-J^i_\nu)\quad\quad (x,y)\mapsto(x'(x),y'(x,y)) \end{cases} where $J^\nu_\mu=\partial_{x^\mu}(x'^\nu(x)), J^i_j=\partial_{y^j}y'^i(x,y),J^i_\nu=\partial_{x^\nu}y'^i(x,y)$ and $\bar{J}^\nu_\mu=(J^\nu_\mu)^{-1}$
Now in the case $E$ is a vector bundle, it makes sense to speak of
A linear Ehresmann connection on a vector bundle $E$ is a connection whose coefficients, in any system of fibred coordinates $(x,y)$ are linear in the fiber coordinates, i.e. $$\Gamma^i_\alpha(x,y)=\Gamma^i_{j\alpha}(x)y^j$$.
Now, I would like to see that this notion of linear connection is equivalent to the one I am used to, the Koszul definition of connection:
Let $\pi:E\rightarrow M$ be a vector bundle. A linear connection on $E$ is a mapping: $$\nabla:\chi(M)\times\Gamma(E)\rightarrow\Gamma(E)\quad (X,V)\mapsto \nabla_XV$$ such that:
1 $\nabla_XV$ is $C^\infty(M)$-linear in $X$, i.e. $$\nabla_{f_1X_1+f_2X_2}V=f_1\nabla_{X_1}V+f_2\nabla_{X_2}V\quad\quad\forall X_1,X_2\in C^\infty(M)\forall V\in\Gamma(E)$$
2 $\nabla_XV$ is $\mathbb{R}$-linear in $V$, i.e. $$\nabla_{X}(a_1V_1+a_2V_2)=a_1\nabla_XV_1+a_2\nabla_XV_2\quad\quad\forall a_1,a_2\in\mathbb{R}\forall V_1,V_2\in\Gamma(E)$$
3 $\nabla$ satisfies the Leibniz rule, i.e. $$\nabla_X(fV)=f\nabla_XV+(Xf)V$$
How can one proceed to do so? Is the operator I asked about in Eheresmann connection and covariant derivative involved?