$\mathbb{R}^4 \times S^1$ is of course a principal $U(1)$-bundle and trivial.
However, I cannot find an explicit example of the Ehresmann connection one-form on this bundle in terms of the coordinates $(x_1, x_2, x_3, x_4, e^{i\theta})$.
Could anyone please provide me a formula for me?
Reading the comments above, I'm still not sure what exactly you are asking for, but I'll try anyway.
A connection on a fiber bundle $\pi:E\to M$ is for each point $p\in E$ a selection of a subspace $Th_pE\subset T_pE$ of the tangent space in that point, which should depend smoothly on $p$ (in a commensurable sense) and which is horizontal (i.e. $Th_pE\oplus\text{ker}\,T\pi\big|_p=T_pE$). Note how the vertical tangent space $\ker\,T\pi\big|_p$ at any $p$ is already determined by the bundle structure. This is, however, not true for a horizontal space (unless in, maybe, some useless, pathological counterexamples...), of course, the space $\text{ker}\,T\pi\big|_p$ has many complements (in the $\oplus$-sense). Further note how, without choosing such a horizontal space, you have no unique projection onto the vertical tangent space, so without prescribing along which (horizontal) complement you want to project, you cannot tell how an arbitrary tangent vector is decomposed into vertical and horizontal component. So conversely, if you choose a projection whose image is $\ker\,T\pi\big|_p$, you have implicitly chosen a horizontal space, namely the kernel of this projection.
This last observation leads to the convenient way to choose a horizontal space in each point, namely by choosing it the kernel of a section in the endomorphism bundle, say $A\in\Gamma^\infty(\text{End}\,E)$. To guarantee the above properties, this section should be a projection, meaning for each $p\in E$ we have $A\big|_p\circ A\big|_p=A\big|_p:T_pE\to T_pE$, whose image in each point $p$ is $\ker\,T\pi\big|_p$, meaning $A\big|_p(v)=v\Leftrightarrow T\pi\big|_p(v)=0$ for all $v\in T_pE$. The "smoothness in a commensurable sense" is now just the smoothness of this section. Any such $A$ is called (Ehresmann) connection (form) on $E$.
So far on the general case. If you now have a trivial bundle $p_M:E=M\times F\to M$, you do have a distinguished horizontal subspace in each point, and thus a distinguished connection form. Therefore note that, in a trivial bundle you don't only have the basis projection $p_M:M\times F\to M$, but also the "fiber projection" $p_F:M\times F\to F$. At some $(x,f)\in M\times F$, the tangent map $p_F\big|_{(x,f)}$ has a kernel, and this kernel is a perfectly fine example of a horizontal space in the above sense. Moreover, this kernel is maybe the most obvious choice of subspace of $T_{(x,f)}(M\times F)$ that should be called "horizontal" or "parallel to $M$" in the present setting of a trivial bundle. This connection is called the trivial connection.
To explicitly construct the connection form which distinguishes this particular choice in each point, note that for each $(x,f)\in M\times F$ the space $T_{(x,f)}(M\times F)$ is canonically isomorphic to $T_xM\oplus T_fF$. Is that clear to you? So, if you mean this connection (which is the only distinguished one in your setting), then you seek for a section $A\in\Gamma\big(\text{End}(T(M\times F))\big)$ which at each point $(x,f)$ is the projection on $T_xM$ along $T_fF$ on the space $T_xM\oplus T_fF$. Can you work out the explicit example you are asking for on your own, starting from here?
Another note on why THE Ehresmann connection is not a thing. In principal bundles you usually formulate the connection form a bit different, but however, do nothing else but distinguishing a horizontal subspace in each tangent space as the pointwise kernel of some differential form on $TE$. Moreover, it is claimed that this collection of horizontal spaces is compatible with the $G$-action, and such connection are then called $G$-connection. So every $G$-connection on a principal bundle $P$ defines an Ehresmann connection on $P$ by merely a bit reformulation, which then does not rely on the principal $G$-bundle structure of $P$ anymore, but only on the fiber bundle structure. But conversely, not every Ehresmann connection on $P$ defines a $G$-connection, as mentioned, therefore it needs an additional property. So I first assumed you to ask for a $G$-connection, where there are many of them, and you specified into the Ehresmann connection, which gives rise to even more possibilities.
I hope, I'm able to help you with some confusion.