We have : $u^Tu = I$, $u$ : is a orthonormal column vector and I = 1$; we differentiate both sides of the formula we get :
$$ (u')^Tu + u^Tu'= 0$$
From derivative of dot product https://en.wikipedia.org/wiki/Vector-valued_function#cite_note-dynon19-1.
Is this correct, it seems to me that this is wrong because $(u')^Tu$ is equal to $u^Tu'$.
No, there are several problems here.
Firstly, if you use transpose notation, you can't swap the order of the elements. $a^Tb$ is the number $a_1 b_1+a_2 b_2+a_3 b_3+ \dotsb$, whereas $ba^T$ is the matrix $$ \begin{pmatrix} b_1 a_1 & b_1 a_2 & b_1 a_3 & \cdots \\ b_2 a_1 & b_2 a_2 & b_3 a_3 & \cdots \\ b_3 a_1 & b_3 a_2 & b_3 a_3 & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} $$
Secondly, one therefore sees from this that while $ u^Tu' = u'^Tu $, $u'u^T \neq uu'^T$: in fact, $u'u^T = (uu'^T)^T$.
On the other hand, you are correct that differentiating $u^Tu = 1$ gives two terms that are the same: namely $$ 0=(u^Tu)' = u'^T u + u^T u' = 2u^T u'. $$ Thus the derivative of $u$ is orthogonal to $u$: this makes sense, since if $u$ changed in the parallel direction, the length of $u$ would change, which is prevented by $u^Tu=1$.