Looking at the coordinate-free expressions that you've found in Wikipedia, it is easy to convince yourself that $\mathrm{grad}$, $\mathrm{curl}$, and $\mathrm{div}$ are instances of the exterior derivative in disguise: just think of the musical isomorphisms and the Hodge star as of means of identification. In the same article one can find a coordinate-free formula, that can be taken as the definition of the exterior derivative. This observation essentially closes the question.
The classical vector calculus deals with $\mathbb{R}^3$, which possesses some specific or exceptional structures, in particular, it has a canonical (Euclidean) coordinate system, the Euclidean metric, and the cross product, that all are extensively used in theory and calculations. If you want tor restict yourself to this case, then I doubt that it is ever possible to find a pure coordinate-free way of expressing the quantities under consideration (i.e. $\nabla f$, $\nabla \cdot \vec{F}$, and $\nabla \times\vec{F}$), as the space $\mathbb{R}^3$ itself is defined by explicitly presenting a single coordinate chart! In other words, you are forced to deal with coordinates and the dimension-related tricks in order to handle these quantities.
Coming back to the expressions in Wikipedia, notice that they use the Hodge star, but we have not received yet any convincing answer on how to give a coordinate-free definition for it. This doubles my pessimism, but I can be wrong and overlook something important.
Nevertheless, I find that this question and the other attempts to answer it are very insightful. For further discussion I suggest to look at the references below.
The best picture that shows that $\mathrm{grad}$, $\mathrm{curl}$, and $\mathrm{div}$ are closely related is given in [1], where they are combined into the de Rham complex. This text is perhaps too advanced, but a diligent undergraduate should be able to follow the first two paragraphs there, and the details can be recovered from [2] and [3].
References:
- M.G. Eastwood, A complex from linear elasticity, http://calvino.polito.it/~salamon/seminar/srni99.pdf
- W.G. Faris,Vector fields and differential forms, September 25, 2008, http://math.arizona.edu/~faris/methodsweb/manifold.pdf
- E.H.Goins, T.M. Washington, A Tasty Combination: Multivariable Calculus and Differential Forms, https://arxiv.org/abs/0910.0047
"Nabla" is a symbolic "vector differential operator". It can be written, symbolically, $\nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial Y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$.
Just as there are three kinds of vector products, "scalar product", which multiplies a vector and a scalar to get a vector, "dot product", which multiplies two vectors to get a scalar, and "cross product", which multiplies two vectors to get a vector, so there are three kinds of "products" with nabla.
"grad" (gradient) which applies to a scalar valued function, f(x, y, z), to get a vector function:
$\nabla f= (\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k})f(x,y,z)= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$"
"div" (divergence) which applies to a vector valued function, $\vec{F(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ to get a scalar valued function: $\nabla\cdot \vec{F}= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$.
"curl" which applies to a vector valued function, $\vec{F(x,y,z)}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, to get a vector valued function:
$\nabla\times F= $$\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z)\end{array}\right|$
$= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$.
Best Answer
We may think of $ \nabla $ as an operator ( del operator ) in the following sense.
It takes a function $f$ and turns it into a vector $\nabla f$ .
$\nabla f= \left\langle \frac {\partial f}{\partial x},\frac {\partial f}{\partial y}, \frac {\partial f}{\partial z} \right\rangle $ is called the gradient vector.
The gradient vector points to the direction at which your function increases most rapidly.
For example if $$ f(x,y,z)= x+3y^2 -10z$$ Then $$ \nabla f (x,y,z)= \langle 1,6y,-10\rangle $$
and if there is a point given, say $(1,3,5)$, we can evaluate $ \nabla f (1,3,5)= \langle 1,18,-10\rangle.$
This vector points at the direction of maximum increase of our function at $(1,3,5).$