= Solution
One definition of a <principal bundle> $P\to M$ uses local trivializations $\phi_i:\pi^{-1}(U_i)\to U_i\times G$ whose overlap maps are
$$
\phi_j\phi_i^{-1}(x,g)=(x,t_{ji}(x)g),
$$
with smooth $G$-valued <transition functions of a principal bundle> satisfying the cocycle condition. This definition displays the global right <group action> rather than assuming it: if $\phi_i(p)=(x,g_i)$, define $\phi_i(p\cdot h)=(x,g_i h)$. On an overlap,
$$
t_{ji}(x)(g_i h)=(t_{ji}(x)g_i)h,
$$
so the local definitions agree. The resulting action is smooth, free and transitive on each fibre, and respects the projection.
If $s:M\to P$ is a global <section of a fiber bundle>, set $F(x,h)=s(x)\cdot h$. The free and transitive fibre action makes $F:M\times G\to P$ bijective. In a local trivialization, write $s(x)=(x,a_i(x))$; then $F(x,h)=(x,a_i(x)h)$ and its inverse is $(x,g)\mapsto(x,a_i(x)^{-1}g)$. Both are smooth and respect the action and projection. Conversely a <principal bundle> isomorphism $M\times G\to P$ sends $x\mapsto(x,e)$ to a global section. Hence
$$
\boxed{P\text{ admits a global section}\ \Longleftrightarrow\ P\cong M\times G\text{ as a principal bundle}.}
$$
For a <pseudo-Riemannian manifold> of <metric signature> $(p,q)$, its <orthonormal frame bundle> has fibre at $x$ consisting of linear isometries $u:\mathbb R^{p,q}\to T_xM$. Its structure group is $O(p,q)$ and its right action is $u\cdot h=u\circ h$. Two such frames differ by a unique element of $O(p,q)$, proving the required free and transitive fibre action. A global section is exactly a global <pseudo-orthonormal frame>. Orientation and time orientation, when specified, restrict the structure group and the allowed frames.
Realize unit-radius four-dimensional <de Sitter spacetime> as
$$
\mathrm{dS}_4=\{X\in\mathbb R^{4,1}:\eta(X,X)=1\},\qquad \eta=\operatorname{diag}(-1,1,1,1,1).
$$
Its tangent space is $X^\perp$, with Lorentzian <metric signature> $(3,1)$. If $e_0,e_1,e_2,e_3$ is a <pseudo-orthonormal frame> at $X$, append $X$ as the final column. The resulting five-by-five matrix preserves $\eta$, and conversely every such matrix supplies a point $X$ and a frame at that point. Thus the full frame bundle is $O(4,1)$, with projection $g\mapsto ge_4$ and fibre group $O(3,1)$.
The printed $SO(4,1)$ statement requires the orientation restriction. Choose the tangent orientation by requiring the appended matrix to have determinant one. Then
$$
\boxed{SO(4,1)\longrightarrow\mathrm{dS}_4,\qquad g\longmapsto ge_4,}
$$
is precisely the <oriented frame bundle of de Sitter spacetime>, with right structure group $SO(3,1)$ embedded as $\operatorname{diag}(h,1)$. If future time orientation is also required, use $SO_0(4,1)$ and $SO_0(3,1)$. Calling $SO(4,1)$ the full bundle without specifying orientation would exclude valid frames of the opposite orientation.
Finally write the global coordinates
$$
X(\tau,n)=(\sinh\tau,\cosh\tau\,n),\qquad \tau\in\mathbb R,\quad n\in S^3.
$$
Direct differentiation gives the <de Sitter spacetime> metric $-d\tau^2+\cosh^2\tau\,g_{S^3}$. The assumed global <frame of a vector bundle> on $S^3$ can be orthonormalized smoothly by <Gram-Schmidt orthogonalization> for the round positive metric. More explicitly, view $S^3$ as the <unit quaternions>: $Y_1(q)=qi$, $Y_2(q)=qj$, $Y_3(q)=qk$ form a smooth global orthonormal tangent basis. Set
$$
E_0=\partial_\tau,\qquad E_i=(\cosh\tau)^{-1}Y_i,\quad i=1,2,3.
$$
These have inner products $g(E_0,E_0)=-1$, $g(E_i,E_j)=\delta_{ij}$ and $g(E_0,E_i)=0$. They are smooth everywhere because $\cosh\tau$ never vanishes. Choose the spatial orientation consistently, reversing one $Y_i$ if necessary, and choose $E_0$ future-directed. This <global frame on four-dimensional de Sitter spacetime> yields a global section of the full frame bundle and also of either restricted bundle. \b[Yes: the de Sitter frame bundle admits a global section and is trivial.]
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