Vector bundles are glued from transition cocycles

The cocycle with $k=i$ gives $g_{ij}{g}_{ji}=I$. Hence the displayed relation is reflexive, symmetric, and transitive: the transitive calculation is $g_{kj}(g_{ji}v)=g_{ki}v$. It therefore defines a quotient $q:{\coprod }_i{U}_i\times {\mathbb{F}}^n\to E$ and a map $p:E\to X$ by $p[x,v,i]=x$. Since $pq(x,v,i)=x$ is continuous on every summand, the quotient property [F2] makes $p$ continuous.

The quotient map $q$ is open. Indeed, if $O$ is open in the coproduct, then the part of its saturation in the $j$th summand is the union over $k$ of the images of $O\cap ((U_j\cap U_k)\times {\mathbb{F}}^n\times \{k\})$ under the homeomorphism $(x,v,k)\mapsto (x,g_{jk}(x)v,j)$; its inverse uses $g_{kj}=g_{jk}^{-1}$. Hence every such part is open. The restriction of $q$ over the saturated open set $p^{-1}(U_i)$ is therefore again a quotient map. Define ${\mathrm{\Phi }}_i[x,v,k]=(x,g_{ik}(x)v)$. The cocycle makes this independent of the representative, and its composite with the restricted quotient map is continuous on every summand, so [F2] makes ${\mathrm{\Phi }}_i$ continuous. Its inverse is $(x,w)\mapsto [x,w,i]$ and is continuous as the $i$th-summand inclusion followed by $q$. Thus ${\mathrm{\Phi }}_i:p^{-1}(U_i)\cong U_i\times {\mathbb{F}}^n$ is a fiberwise-linear chart. Its overlap from $i$ to $j$ is $g_{ji}$, so [F1] proves that $E$ is the claimed bundle.

For the primed cocycle, the maps on summands $(x,v,i)\mapsto [x,h_i(x)v,i{]}^{\mathrm{\prime }}$ respect the relation because $g_{ji}^{\mathrm{\prime }}{h}_i=h_j{g}_{ji}$. By [F2] they descend to a continuous fiberwise-linear map $E\to E^{\mathrm{\prime }}$. Replacing $h_i$ by $h_i^{-1}$ gives its continuous inverse, so it is a bundle isomorphism.

Finally, a linear atlas of a rank-$n$ bundle supplies the functions $g_{ji}$ and their cocycle law by [F1]. Sending the quotient class $[x,v,i]$ to ${\varphi }_i^{-1}(x,v)$ is well-defined, continuous by [F2], and in each chart is the identity map on $U_i\times {\mathbb{F}}^n$. It is therefore a bundle isomorphism from the reconstructed quotient to the original bundle.