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 orthonormal frames of $g$, 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 $1$-forms $\omega_{ij}=-\omega_{ji}$ on $B$ such that $\mathrm{d}\omega_i = - \omega_{ij}\wedge\omega_j$. (NB: My $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. I'm sorry about that, but I just can't switch back to Chern's conventions without getting confused.) The curvature $2$-forms are $\Omega_{ij} = \mathrm{d}\omega_{ij} + \omega_{ik}\wedge\omega_{kj} = \tfrac12 R_{ijkl}\,\omega^k\wedge\omega^l$, 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 an $n$-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 $\Omega$. Chern denotes them both by the same letter $\Omega$. 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'. What he does is let $u = (u_i):S^{n-1}\to \mathbb{R}^n$ 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 $M$ 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.

There's actually a way to avoid 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 $\mathrm{SO}(n{-}1)$-orbits that are the leaves of the system $$ \omega_1=\omega_2=\cdots=\omega_n=\omega_{12}=\omega_{13}=\cdots=\omega_{1n}=0. $$ Then you can construct an $(n{-}1)$-form $\tilde\Pi$ directly on $B$ out of these forms that will satisfy $\mathrm{d}\tilde\Pi = \tilde\Omega$. This $\tilde \Pi$ will be the $\pi_1$-pullback of a 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 $n$.

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.