Short answer
You can calculate the normal derivative of a vector, it is given by the projection of the gradient of the vector (a second-order tensor) in the normal direction. In cartesian coordinates, this is
$$ \nabla_n u = \nabla u \cdot n = \left[\begin{matrix}\frac{\partial}{\partial x} u_x & \frac{\partial}{\partial y} u_x & \frac{\partial}{\partial z} u_x\\
\frac{\partial}{\partial x} u_y & \frac{\partial}{\partial y} u_y & \frac{\partial}{\partial z} u_y\\
\frac{\partial}{\partial x} u_z & \frac{\partial}{\partial y} u_z & \frac{\partial}{\partial z} u_z\end{matrix}\right]
\left[\begin{matrix}n_{x}\\n_{y}\\n_{z}\end{matrix}\right] =
\left[\begin{matrix}n_{x} \frac{\partial}{\partial x} u_x + n_{y} \frac{\partial}{\partial y} u_x + n_{z} \frac{\partial}{\partial z} u_x\\n_{x} \frac{\partial}{\partial x} u_y + n_{y} \frac{\partial}{\partial y} u_y + n_{z} \frac{\partial}{\partial z} u_y\\n_{x} \frac{\partial}{\partial x} u_z + n_{y} \frac{\partial}{\partial y} u_z + n_{z} \frac{\partial}{\partial z} u_z\end{matrix}\right]
$$
Long answer
In linear elasticity, the constitutive relation (Hooke's law) relates the strain (a second-order tensor) to the stress (a second-order tensor) by the stiffness tensor (a fourth-order tensor), i.e.
$$\sigma = C \varepsilon\, ,$$
or, in index notation,
$$\sigma_{ij} = C_{ijkl} \varepsilon_{kl}\, ,$$
where summation over repeated indices is implied. The stiffness tensor has the following symmetries $C_{ijkl} = C_{ijlk} = C_{jikl} = C_{klij}$.
Furthermore, the strain tensor is given by the symmetric part of the gradient of the displacement vector:
$$\varepsilon = \frac{1}{2}\left[\nabla u + (\nabla u)^T\right]\, ,$$
or, in index notation,
$$\varepsilon = \frac{1}{2}(u_{i,j} + u_{j, i})\, ,$$
where the comma represents a spatial derivative.
Thus, due to symmetries in the stiffness tensor, we can say that
$$\sigma = C \varepsilon = C\nabla u\, .$$
But, that does not imply that we can invert the relation to directly obtain the gradient of $u$ from $\sigma$.
This is why:
$$\sigma = C\nabla u = C \varepsilon + \underbrace{C\omega}_{=0} \, ,$$
thus,
$$S\sigma = SC \varepsilon = \varepsilon \, ,$$
where $S$ is the compliance tensor, and $S_{ijkl}C_{klrs}=\frac{1}{2}(\delta_{ir}\delta_{js} + \delta_{is}\delta_{jr})$.