In the Magnus and Neudecker's book "Matrix differential calculus with applications", Sec. 8.9, the authors derive an expression for the derivative of the right eigenvector $u(t)$ corresponding to a simple eigenvalue $\lambda(t)$ of a matrix $Z(t)\in M_n(\mathbb C)$: $$u'(0)=(\lambda I -Z(0))^\dagger \left(I-\frac{u(0)v^*(0)}{v^*(0)u(0)}\right)Z'(0)u(0),$$ where $()^\dagger$ is the Moore-Penrose inversion operator and $v^T(t)$ is the left eigenvector.
When deriving this expression they normalize $u(t)$ s.t. $\langle u(t),u(0)\rangle=1$ for small $t$, thus basically requiring that $\langle u'(0),u(0)\rangle=0$.
So my questions are:
Is it always possible to perform such a normalization?
How would you do this in practice when you do not know $u(t)$, which is obviously the case as otherwise you wouldn't need to use the above expression?
Sure, it's possible to perform this normalization except in the case that the right eigenvector corresponding to eigenvalue $\lambda$ becomes orthogonal to $u(0)$, which is presumably why they consider only small $t$.
Think about it this way: the eigenvector corresponding to a given eigenvalue is only defined up to a complex constant, and it is not clear how to differentiate $u(t)$ if you allow its image to be an entire one-dimensional subspace, rather than a vector. So instead of considering all possible rescalings of the eigenvector, Magnus and Neudecker are selecting one particular representative element in the equivalence class of vectors which have $\lambda$ as an eigenvalue; this representative element sweeps out a curve in $\mathbb{C}^n$ as $t$ varies and can be differentiated.
In the real case, the natural normalization is to require $\|u(t)\|^2=1$, but obviously this does not work in the complex domain.
Finally I'm not sure what the "this" is you're asking about in the second part of the question: the formula you've listed only requires knowing $u(0$), not $u(t)$ for other $t$, but in any case performing the normalization is easy: if $v$ is an eigenvector with eigenvalue $\lambda$ at some $t$, you normalize it by computing $$u(t) = \frac{v}{\langle v, u(0)\rangle}.$$