The key idea is to use Y-rotations, because they include the imaginary unit which cancels the imaginary unit present in the exponential formula.
Start with $U_2$. Note that it is an orthogonal matrix, because $e, f, g, h \in \mathbb{R}$. Therefore there exists angle $\theta$ such that
$$
U_2(\theta) = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \cos \theta/2 & -\sin \theta/2 \\
0 & 0 & \sin \theta/2 & \cos \theta/2
\end{pmatrix}
$$
or
$$
U_2(\theta) = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \cos \theta/2 & \sin \theta/2 \\
0 & 0 & \sin \theta/2 & -\cos \theta/2
\end{pmatrix}.
$$
These two forms correspond to the two connected components of $O(2)$, i.e. to a rotation and a rotoreflection, respectively. Noting that
$$
R_y(\theta) = \exp(-i\theta Y/2) = I \cos \frac{\theta}{2} - i Y \sin \frac{\theta}{2} = \begin{pmatrix}
\cos \theta/2 & -\sin \theta/2 \\
\sin \theta/2 & \cos \theta/2
\end{pmatrix}
$$
we see that in the rotation case
$$
U_2(\theta) = C^{(1)}R_y^{(2)}(\theta)
$$
and in the rotoreflection case
$$
U_2(\theta) = C^{(1)}R_y^{(2)}(\theta) \, C^{(1)}Z^{(2)}
$$
where $C^{(i)}$ indicates that the $i$th qubit is the control and $R_y^{(j)}$ and $Z^{(j)}$ indicate that the $j$th qubit is the target.
We can extend the above result to $U_1$ using the $CNOT$ gate. Specifically, we exploit the fact that the matrix of the $C^{(2)}NOT^{(1)}$ gate is the permutation matrix that swaps the second and fourth basis elements
$$
C^{(2)}NOT^{(1)} = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0
\end{pmatrix}.
$$
As before, we notice that $U_1$ is orthogonal due to the requirement that $a, b, c, d \in \mathbb{R}$ and therefore there exists angle $\theta$ such that
$$
U_1(\theta) = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos \theta/2 & -\sin \theta/2 & 0 \\
0 & \sin \theta/2 & \cos \theta/2 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
or
$$
U_1(\theta) = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos \theta/2 & \sin \theta/2 & 0 \\
0 & \sin \theta/2 & -\cos \theta/2 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
where once again the two forms correspond to a rotation and a rotoreflection, respectively. We reduce the rotation case of $U_1$ to the rotation case of $U_2$ using
$$
U_1(\theta) = C^{(2)}NOT^{(1)} \, U_2(-\theta) \, C^{(2)}NOT^{(1)}
$$
and the rotoreflection case of $U_1$ to the rotoreflection case of $U_2$ using
$$
U_1(\theta) = C^{(2)}NOT^{(1)} \, U_2(-\theta) \, Z^{(1)} \, C^{(2)}NOT^{(1)}
$$
where $Z^{(1)}$ is the single-qubit $Z$ gate applied to the first qubit.