The definition is that $A\in M_{n}(\mathbb{R})$ is called a rotation
matrix if there exist a unitary matrix $P$ s.t $P^{-1}AP$ is of
the form $$\begin{pmatrix}\cos(\theta) &-\sin(\theta)\\
\sin(\theta) & \cos(\theta)\\
& & 1\\
& & & 1\\
& & & & 1\\
& & & & & .\\
& & & & & & .\\
& & & & & & & .\\
& & & & & & & & 1
\end{pmatrix}$$
If we consider $A:\mathbb{R}^{n}\to\mathbb{R}^{n}$ then the meaning
is that there exist an orthonormal basis where we rotate the $2-$dimensional
space spanned by the first two vectors by angle $\theta$ and we fix
the other $n-2$ dimensions
Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ...
$$\begin{align}
T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\
R(\mathbf{x}) &= (2c^2-1)\mathbf{x} + 2 s^2 (\mathbf{x}\cdot\mathbf{n}) \mathbf{n} + 2 s c (\mathbf{x}\times\mathbf{n}) \\
&= (2c^2-1)\mathbf{x} + 2 (\mathbf{x}\cdot\mathbf{w})\mathbf{w} + 2 c (\mathbf{x}\times\mathbf{w}) \\
\end{align}$$
Decomposing $\mathbf{x}$ as $p\mathbf{u} + q\mathbf{v} + r \mathbf{w}$, we can get fairly directly ...
$$\mathbf{x}\cdot\mathbf{u} = p + q c \qquad \mathbf{x}\cdot\mathbf{v}=pc+q \qquad \mathbf{x}\cdot\mathbf{w}=rs^2 \qquad (\star)$$
$$\mathbf{x}\times\mathbf{w} = \mathbf{x}\times \left(\mathbf{u}\times\mathbf{v}\right) = (\mathbf{x}\cdot\mathbf{v})\mathbf{u}-(\mathbf{x}\cdot\mathbf{u})\mathbf{v} \qquad (\star\star)$$
Then it's straightforward to show that the difference of the transformations vanishes:
$$\begin{align}
T(\mathbf{x}) - R(\mathbf{x}) &=\mathbf{x}-2(p+qc)\mathbf{u}-2(pc+q)\mathbf{v} + 4c(pc+q)\mathbf{u}\\
&-\left( (2c^2-1)\mathbf{x} + 2 r s^2 \mathbf{w} + 2 c \left( (pc+q)\mathbf{u} - (p+qc)\mathbf{v} \right) \right) \\[6pt]
&= (2-2c^2)\;\mathbf{x} + 2 \left(-p-qc+2pc^2+2qc-pc^2-qc\right)\;\mathbf{u} \\
&+ 2\left(-pc-q+pc+qc^2\right)\mathbf{v} - 2 r s^2 \mathbf{w} \\[6pt]
&= 2 s^2 \left( \mathbf{x} - p\mathbf{u} - q \mathbf{v} - r\mathbf{w} \right) \\[6pt]
&= 0
\end{align}$$
and we conclude that the transformations are equivalent. $\square$
Edit. Without jumping immediately to the decomposition of $\mathbf{x}$, we can use the expansion in $(\star\star)$ to write
$$\begin{align}
\frac{T(\mathbf{x})-R(\mathbf{x})}{2s^2} \;\;&=\;\; \mathbf{x}
\;-\; \left( \; \frac{\mathbf{x}.( \mathbf{u} - c \mathbf{v} )}{s^2}\;\mathbf{u}
\;+\; \frac{\mathbf{x}.( \mathbf{v} - c \mathbf{u} )}{s^2}\;\mathbf{v}
\;+\; \frac{\mathbf{x}.\mathbf{w}}{s^2}\;\mathbf{w} \;\right)
\end{align}$$
If you can "see" that the coefficients of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are the components of $\mathbf{x}$ ---which would be clear for orthogonal $\mathbf{u}$ and $\mathbf{v}$, for which $c=0$ and $s=1$--- then you're done. If not, note that you can arrive at this insight by solving the dot-product equations $(\star)$ for $p$, $q$, $r$.
Best Answer
You don’t need to compute the angle explicitly, or indeed refer to an angle at all.
Observe that the result of rotating any vector $(x,y)^T$ 90 degrees counterclockwise is $(-y,x)^T$. Then, using the fact that the columns of a transformation matrix are the images of the basis vectors, the matrix $$\begin{bmatrix}x&-y\\y&x\end{bmatrix}$$ represents a rotation that maps $(1,0)^T$ onto the unit vector $(x,y)^T$. Therefore, a rotation that takes the unit vector $\mathbf a=(x_a,y_a)^T$ to the unit vector $\mathbf b=(x_b,y_b)^T$ has the matrix $$\begin{bmatrix}x_b&-y_b\\y_b&x_b\end{bmatrix} \begin{bmatrix}x_a&-y_a\\y_a&x_a\end{bmatrix}^{-1} = \begin{bmatrix}x_b&-y_b\\y_b&x_b\end{bmatrix} \begin{bmatrix}x_a&y_a\\-y_a&x_a\end{bmatrix} = \begin{bmatrix}x_ax_b+y_ay_b&x_by_a-x_ay_b \\ x_ay_b-x_by_a & x_ax_b+y_ay_b\end{bmatrix}.$$