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Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also obtain this as follows. For every $p\in P$, you can define ${d\over{dt}}exp(tu).p\in T_pP$ and check that locally it corresponds to the previous definition.

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also define: For every $p\in P$, you can define ${d\over{dt}}exp(tu).p\in T_pP$.

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also obtain this as follows. For every $p\in P$, you can define ${d\over{dt}}exp(tu).u\in T_pP$ and check that locally it corresponds to the previous definition.

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also obtain this as follows. For every $p\in P$, you can define ${d\over{dt}}exp(tu).p\in T_pP$ and check that locally it corresponds to the previous definition.

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a left invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also obtain this as follows. For every $p\in P$, you can define ${d\over{dt}}exp(tu).u\in T_pP$ and check that locally it corresponds to the previous definition.

Suppose that the bundle $f:P\rightarrow M$ is defined by the trivialisation $(U_i,\phi_i, g_{ij})$ $\phi_i:p^{-1}(U_i)\rightarrow U_i\times G$. You have $p^{-1}U_i)=U_i\times G$. Let $g$ be the Lie algebra of $G$. For each element $u\in g$ correspond to a right invariant vector defined on $G$. That is you identify $u$ with an element of $T_IG$ the tangent space of $G$ at the identity $I$. And for $g\in G$, $u(g)=dR_g(u)$ where $R_g$ is the right multiplication by $g$. Locally for each $(x,g)\in p^{-1}(U_i)=U_i\times G$ you can define $X^i_u(x,g)=(0,u(g)$. This definition does not depend of the trivialisation:

Suppose that $f(p)\in U_i\cap U_j$, write $\phi_j(p)=(x,g)$, you then have $\phi_i(p)=(x,gg_{ij}(x))$. This implies that $X^i(g)=(0,dR_{gg_{ij}}(u)=d((R_{g_{ij}}R_g)(u)= dRg_{ij}X^j(u)$.

You can also obtain this as follows. For every $p\in P$, you can define ${d\over{dt}}exp(tu).u\in T_pP$ and check that locally it corresponds to the previous definition.