It would be misleading to write $$\vec P=(r, \theta, \phi)$$ or $$\vec P=\begin{pmatrix}r \\ \theta \\ \phi \end{pmatrix}$$ because the curvilinear coordinates $r,\theta,\phi$ don't behave like vector components with the usual simple rules for vector addition and multiplication. And it would easily lead to nonsensical conclusions (as you already noticed).
It is better to imagine $\vec{P}$ with cartesian components $x,y,z$ because these truly behave like vector components. $$\vec{P} =\begin{pmatrix}x \\ y \\ z\end{pmatrix} =\begin{pmatrix}r\cos\theta\cos\phi \\ r\cos\theta\sin\phi \\ r\sin\theta \end{pmatrix} \tag{1}$$ $$\vec{P}' =\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} =\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$
With this notation you get the vectors $\vec e_\theta,\vec e_\phi,\vec e_r$ in a straight-forward way by differentiating (1).
$$\vec{e}_r = \frac{\partial\vec P}{\partial r} =\begin{pmatrix}\cos\theta\cos\phi \\ \cos\theta\sin\phi \\ \sin\theta \end{pmatrix}$$ $$\vec{e}_\theta = \frac{\partial\vec P}{\partial \theta} =\begin{pmatrix}-r\sin\theta\cos\phi \\ -r\sin\theta\sin\phi \\ r\cos\theta \end{pmatrix}$$ $$\vec{e}_\phi = \frac{\partial\vec P}{\partial \phi} =\begin{pmatrix}-r\cos\theta\sin\phi \\ r\cos\theta\cos\phi \\ 0 \end{pmatrix}$$
It is meant with the vectors above that you have equation $$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$