Intuitive proof for the second directional derivative

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I understand that the second directional derivative can be found by taking the gradient vector of the first directional derivative and multiplying it by the given unit vector, as follows:

$$\nabla^2_{\hat{u}} f(x,y,z) := \left\langle\frac{\partial\nabla_{\hat{u}}f(x,y,z)}{\partial x},\frac{\partial\nabla_{\hat{u}}f(x,y,z)}{\partial y},\frac{\partial\nabla_{\hat{u}}f(x,y,z)}{\partial z}\right\rangle\cdot \hat{u}$$

My question is why the second directional derivative isn't instead defined by the following formula:

$$\nabla^2_{\hat{u}} f(x,y,z) \stackrel{?}{=} \underbrace{\left\langle \frac{\partial^2 f(x,y,z)}{\partial x^2}, \frac{\partial^2 f(x,y,z)}{\partial y^2}, \frac{\partial^2 f(x,y,z)}{\partial z^2}\right\rangle}_{\textrm{The second derivative of the gradient vector}}\cdot \underbrace{\hat{u}}_{\textrm{Unit vector}}$$

Basically why isn't the second directional derivative just the dot product of the second derivative of the gradient vector and the unit vector?