Basic rotations in $3-D$ are represented by the matrices: $$ R^3_{xy}(\theta)= \begin{bmatrix} \cos \theta &-\sin \theta & 0\\ \sin \theta &\cos \theta & 0\\ 0& 0 & 1\\ \end{bmatrix}\quad \text{rotation in the plane}\; (xy) $$
$$ R^3_{xz}(\theta)= \begin{bmatrix} \cos \theta&0 &\sin \theta\\ 0& 1 & 0\\ -\sin \theta & 0&\cos \theta \\ \end{bmatrix}\quad \text{rotation in the plane}\; (xz) $$ $$ R^3_{yz}(\theta)= \begin{bmatrix} 1& 0 & 0\\ 0 &\cos \theta&-\sin \theta\\ 0&\sin \theta & \cos \theta \\ \end{bmatrix}\quad \text{rotation in the plane}\; (yz) $$ where te sign of the $\sin \theta$ components are determined by the right-hand rule.
In $4-D$ space the basic rotations becomes something as $$ R^4_{xy}(\theta)= \begin{bmatrix} \cos \theta &-\sin \theta & 0&0\\ \sin \theta &\cos \theta & 0&0\\ 0& 0 & 1&0\\ 0& 0 & 0&1\\ \end{bmatrix} \quad R^4_{xz}(\theta)= \begin{bmatrix} \cos \theta&0 &\sin \theta&0\\ 0& 1 & 0 &0\\ -\sin \theta & 0&\cos \theta&0 \\ 0&0&0&1 \end{bmatrix} $$
$$ R^4_{xt}(\theta)= \begin{bmatrix} \cos \theta&0&0 &-\sin \theta\\ 0& 1 & 0&0\\ 0& 0 & 1&0\\ \sin \theta&0&0 &\cos \theta \\ \end{bmatrix} \quad R^4_{yz}(\theta)= \begin{bmatrix} 1 & 0 &0 &0\\ 0&\cos \theta&\sin \theta&0\\ 0&-\sin \theta & \cos \theta &0\\ 0&0&0&1 \end{bmatrix} $$
$$ R^4_{yt}(\theta)= \begin{bmatrix} 1& 0& 0&0\\ 0&\cos \theta&0 &\sin \theta\\ 0& 0 & 1&0\\ 0&-\sin \theta&0 &\cos \theta \\ \end{bmatrix} \quad R^4_{zt}(\theta)= \begin{bmatrix} 1 & 0 &0 &0\\ 0&1&0&0\\ 0&0&\cos \theta &-\sin \theta\\ 0&0&\sin \theta &\cos \theta \\ \end{bmatrix} $$ But here (as far as i know) we don't have a simple extension of the right-hand rule, so what is the rule for the signs of the $\sin \theta$ components?
In $4D$ there are ${4\choose 2}=6$ basic rotation matrices, each one characterized by the two out of four axes that they keep fixed while rotating vectors in the plane perpendicular to those axes. Alternatively, we can characterize them by the two out of four basis vectors they rotate. We know from $3D$ that the right-hand rule (counter clockwise rotation) requires a minus sign at exactly one of the $\sin\theta$ elements in the upper right corner of the $2\times 2$ submatrix that does the rotation.
A plus sign at exactly one of those $\sin\theta$ elements will clearly lead to the left-hand rule (clockwise rotation).
(We don't worry about the $\sin\theta$ in the lower left corner as this always must have the opposite sign to make the matrix orthogonal.)
The $4D$ matrices in the question satisfy the right or left hand rule for every $3D$ subspace:
\begin{array}{|c|c|c|c|} \hline \text{ subspace}&\text{matrices}&\text{rule}&\text{rotation}\\ \hline xyz&R^4_{xy},R^4_{xz},R^4_{yz}&\text{right hand}&\text{counter clockwise }\\ yzt&R^4_{yz},R^4_{yt},R^4_{zt}&\text{right hand}&\text{counter clockwise }\\ xyt&R^4_{xy},R^4_{xt},R^4_{yt}&\text{left hand}&\text{clockwise}\\ xzt&R^4_{xz},R^4_{xt},R^4_{zt}&\text{left hand}&\text{clockwise}\\ \hline \end{array}
Trying to enforce the right hand rule for every $3D$ subspace seems impossible. For example: change $\sin\theta$ to $-\sin\theta$ in $R^4_{yt}$ would lead to a right hand rule in subspace $xyz$ but break it in the subspace $yzt\,.$