I see that the OP may not be entirely convinced by my comments, so let me try this, which may help. It's understandable that reading the older literature can be confusing; the classical language is often quite different from ours. Here is a slightly different interpretation of Chern's construction of the transgressed form $\Pi$ that may make it clearer what is going on.
Let $n=2p>0$ be an even integer and let $(R^n,g)$ be an oriented Riemannian $n$-manifold. Let $\pi:B\to R$ be the principal right $\mathrm{SO}(n)$-bundle consisting of the oriented $g$-orthonormal frames on $R$, i.e., each $e\in B$ with $\pi(e)=x\in R$ is of the form $e = (e_1,\ldots e_n)$ where $(e_1,\ldots,e_n)$ form an oriented, $g$-orthonormal basis of $T_xR$.
As usual, one has the tautological $1$-forms $\omega_i$ defined by $\omega_i(v) = e_i\cdot \pi'(v)$ for $v\in T_eB$. There exist unique connection $1$-forms $\omega_{ij}=-\omega_{ji}$ on $B$ satisfying the first structure equations, $$\mathrm{d}\omega_i = - \omega_{ij}\wedge\omega_j.$$ (NB: My $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. I'm sorry about that, but I can't switch back to Chern's conventions without getting confused. On the bright side, the first structure equations actually play no role in the sequel, which is why the Gauss-Bonnet formula for the Euler class works for any $\mathrm{SO}(n)$-connection on any oriented orthogonal bundle over $R$, not just the tangent bundle.)
The curvature $2$-forms are defined by the second structure equations
$$\Omega_{ij} = \mathrm{d}\omega_{ij} + \omega_{ik}\wedge\omega_{kj} = \tfrac12 R_{ijkl}\,\omega^k\wedge\omega^l,
$$
and they satisfy the Bianchi identities $\mathrm{d}\Omega_{ij} = \Omega_{ik}\wedge\omega_{kj}-\omega_{ik}\wedge\Omega_{kj}$.
The Gauss-Bonnet $n$-form on $B$ is
$$
\tilde\Omega = \frac1{2^{n}\pi^{n/2}(n/2)!}\ \epsilon_{i_1i_2\cdots i_n}
\Omega_{i_1i_2}\wedge\Omega_{i_3i_4}\wedge\cdots\wedge \Omega_{i_{n-1}i_n}\ ,
$$
where the sum is over all permutations in $S_n$. The Bianchi identities imply that $\mathrm{d}\tilde\Omega=0$, and, since $\tilde\Omega$ is a multiple of $\omega_1\wedge\cdots\wedge\omega_n$, it follows that there exists a unique $n$-form $\bar\Omega$ on $M$ such that $\pi^*\bar\Omega = \tilde\Omega$.
All this is standard, except that, for clarity, I am distinguishing $\tilde\Omega$ and $\bar\Omega$. (Chern uses the same letter $\Omega$ for both.) Similarly, below, I will try to distinguish forms that Chern identifies when they differ via pullback under a submersion with connected fibers.
What Chern does next, though, is what seems to be causing the confusion about 'arbitrary vector fields'. He lets $u = (u_i):S^{n-1}\to \mathbb{R}^n$ denote the inclusion of the $(n{-}1)$-sphere into $\mathbb{R}^n$, and he defines a submersion $\bar\pi: B\times S^{n-1}\to M^{2n-1}$, where $M^{2n-1}\subset TR$ is the unit sphere bundle, by $\bar\pi(e,u) = u_ie_i$. He then constructs an $(n{-}1)$-form $\tilde\Pi$
on $B\times S^{n-1}$ that has the property that $\mathrm{d}\tilde\Pi = \tilde\Omega$ and has the property that there exists an $(n{-}1)$-form $\Pi$ on $M$ (not $R$) satisfying $\bar\pi^*\Pi = \tilde\Pi$. It is important to recognize that Chern's $u_i$ are not the components of a vector field on anything, arbitrary or otherwise.
N.B.: Actually, Chern says that he is going to work locally on an open subset (say) $O\subset R$ and choose a section of $B$ over $U$, i.e., an 'oriented orthonormal frame field' on $O$, he identifies the part of $M$ that lies over $O$ with $O\times S^{n-1}$ and constructs $\Pi$ on $O\times S^{n-1}$, and then he says that the resulting $\Pi$ doesn't depend on the local choice of frame field. However, he never actually uses the local section in any of his calculations, so they are perfectly valid on $B\times S^{n-1}$.
Now, there's a way to avoid introducing the $u_i$ at all, and you may like this better. Consider the submersion $\pi_1:B\to M$ defined by $\pi_1(e) = e_1$. The fibers of this map are (connected) $\mathrm{SO}(n{-}1)$-orbits that are the leaves of the system
$$
\omega_1=\omega_2=\cdots=\omega_n=\omega_{12}=\omega_{13}=\cdots=\omega_{1n}=0.
$$
Then one constructs an $(n{-}1)$-form $\tilde\Pi$ directly on $B$
that is a polynomial with constant coefficients in the forms
$$
\{\omega_{12},\omega_{13},\cdots,\omega_{1n}\}\cup
\bigl\{\ \Omega_{ij}\ \bigl|\ 1<i,j\le n\ \bigr\}
$$
and that will satisfy $\mathrm{d}\tilde\Pi = \tilde\Omega$.
This $\tilde \Pi$ will be the $\pi_1$-pullback of a unique form $\Pi$ on $M$
that does the job.
For example, as Ben remarked, when $n=2$, you can take $$\tilde\Pi = \frac{\omega_{12}}{2\pi},$$ since, in that case, we have
$$
\mathrm{d}\left(\frac{\omega_{12}}{2\pi}\right)
= \frac{\Omega_{12}}{2\pi} = \frac{K\ dA}{2\pi}
$$
When $n=4$, you can take
$$
\tilde\Pi = \frac{2\,\omega_{12}\wedge\omega_{13}\wedge\omega_{14}
+\bigl(\omega_{12}\wedge\Omega_{34}+\omega_{13}\wedge\Omega_{42}
+\omega_{14}\wedge\Omega_{23}\bigr)}{(2\pi)^2}
$$
and verify, using the structure equations and the Bianchi identities, that
$$
\mathrm{d}\tilde\Pi = \frac{
\bigl(\Omega_{12}\wedge\Omega_{34}+\Omega_{13}\wedge\Omega_{42}
+\Omega_{14}\wedge\Omega_{23}\bigr)}{(2\pi)^2} = \tilde\Omega
$$
Of course, Chern computes the explicit formula for all $n$. To put Chern's general formula in the above form, you can just set $u_1=1$, $u_2=u_3=\cdots=u_n=0$ in Chern's formulae (and switch the signs appropriately because my $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. That's all there is to it.
