@JamalS has provided the more formal reason as to why you don't want to think of $\nabla$ as a vector, but here are some reasons (convenient but not formal) as to why it can be useful to do so.
$\vec\nabla\cdot\vec h$ does have some of the meaning of a scalar product in the sense that
$$
\vec a\cdot \vec b= a_xb_x+a_yb_y+a_zb_z
$$
gets copied to
$$
\vec\nabla \cdot \vec h=\frac{\partial}{\partial x}f_x+
\frac{\partial}{\partial y}f_y+\frac{\partial}{\partial z}f_z
$$
if you think of $\vec \nabla=\hat x\frac{\partial}{\partial x}+
\hat y\frac{\partial}{\partial y}+\hat z\frac{\partial}{\partial z}$, i.e. if you think of $\vec\nabla$ as a vector with components $\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$.
Moreover, there are several vector identities that have an vector calculus interpretation if you think of $\vec \nabla$ as a vector. For instance, since $\vec a\times\vec a=0$, you can "export" this to get insight into
$$
\vec\nabla\times \vec \nabla f=0\, ,
$$
for any scalar function $f$. You can also "export" $\vec a\cdot (\vec a\times \vec b)=0$ to intuitively understand
$$
\vec\nabla\cdot \left(\vec\nabla\times \vec A\right)=0\, .
$$
Also a vector multiplied by a scalar yields a vector, much like $\vec \nabla f$ is a vector.
However, not all properties of the usual scalar product can be "exported" to $\vec\nabla$. For instance, $\vec a\cdot \vec b=\vec b\cdot \vec a$ but
$$
(\vec\nabla \cdot \vec h)f\ne (\vec h\cdot\vec\nabla)f\, .
$$
So, while convenient to think of $\vec \nabla\cdot\vec h$ as a scalar product, and convenient to think of $\vec \nabla$ as a vector, one should do so with care else blind manipulations may get you in trouble.