I have vector $w=(0,1)$ and unknow vector $v, w+v=u$, in two dimension coordinate system. After performing rotation $R$ (angle is below 90*) vectors are $w', v', u'$.
If vector $v=0$ calculating $R$ when you know only $u'=w'$ is not a problem $w'=Rw, R=w'w^{-1}$.
But what about the situation, when $v$ is not a zero vector. Is this possible to calculate $R$, when you only know $u'$ which is equal $w'+v'$ and $w$.
I have tons of papers with drawings, but i got nothing. Angle is below 90 degree, so $tg \theta$ is defined, but it seems to be not much helpfull.
Consider the normal 2D rotation matrix: $$\mathbf{R} = \left [ \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right ] = \left [ \begin{matrix} r_c & -r_s \\ r_s & r_c \end{matrix} \right ]$$ If $0° \le \theta \le 90°$, then $0 \le r_c \le 1$ and $0 \le r_s \le 1$.
Because $\mathbf{R}$ is a real orthogonal matrix, we also have $$\begin{cases} r_c \in \mathbb{R} \\ r_s \in \mathbb{R} \\ r_c^2 + r_s^2 = 1 \end{cases}$$ In the first quadrant, $0° \le \theta \le 90°$, we can therefore say that $$r_s = \sqrt{1 - r_c^2}$$ (Note that in general, $r_s = \pm \sqrt{1 - r_c^2}$, and $r_c = \pm \sqrt{1 - r_s^2}$.)
Now, let's say we have vectors $$\begin{array}{l} \vec{w} = ( 0 , 1 ) \\ \vec{p} = ( x_p , y_p ) \\ \vec{v} = ( x_v , y_v ) \end{array}$$ so that $\vec{w}+\vec{p}$ rotated by $\mathbf{R}$ yields $\vec{v}$: $$\mathbf{R}\left(\vec{w} + \vec{p}\right) = \vec{v}$$ i.e. $$\left [ \begin{matrix} r_c & -\sqrt{1-r_c^2} \\ \sqrt{1-r_c^2} & r_c \end{matrix} \right ] \left [ \begin{matrix} x_p \\ y_p + 1 \end{matrix} \right ] = \left [ \begin{matrix} x_v \\ y_v \end{matrix} \right ]$$ The above is essentially a pair of equations, $$\begin{cases} x_v = r_c x_p - \sqrt{1 - r_c^2} ( y_p + 1 ) \\ y_v = \sqrt{1 - r_c^2} x_p + r_c ( y_p + 1 ) \\ \end{cases}$$
If we knew $r_c$, then we could solve $( x_p , y_p )$ from the above: $$\begin{cases} x_p = x_v r_c + y_v \sqrt{1 - r_c^2} \\ y_p = y_v r_c - 1 - x_v \sqrt{1 - r_c^2} \end{cases} \tag{1}\label{1}$$
Oops. That means that you can choose any $r_c \in \mathbf{R}$, $0 \le r_c \le 1$, and there is at least one $\vec{p} = ( x_p , y_p )$ that fulfills the requirements!
In short, $r_c$ and therefore $\mathbf{R}$, cannot be determined if you only know $\vec{w}$ and $\vec{v}$ and nothing about $\vec{p}$.
(In OP's notation, knowing only $w$ and $w' + v'$ and nothing about $v$ is not sufficient to calculate the rotation matrix $R$.)
We need to know more about $\vec{p}$ to define $\mathbf{R}$. Its length (norm) is not sufficient, because we already know it; $\lVert \vec{w} + \vec{p} \rVert = \lVert \vec{v} \rVert$ because $\mathbf{R}$ is a pure rotation matrix, no scaling. You can apply restrictions on $\vec{p}$ using equation $\eqref{1}$ (in inequality form) and solving for $r_c$. The idea is, you need a sufficient restriction that limits $r_c$ to just one value (within $[0 , 1]$).