TITLE: What is the intuition behind the definition of the differential of a function?
QUESTION [13 upvotes]: What is the intuition behind the definition of a differential of a function in differential geometry? i.e. $$df(p)(v_{p}) =v_{p} (f)(p) $$ where $v_{p} \in T_{p} M$ is a vector in the tangent space to the point $p\ M$. 
Is it just that the notion of an infinitesimal change in a function is not mathematically rigorous (in the "traditional" sense that it's presented in elementary calculus). So in order to make the notion mathematically rigorous we note that a quantity that captures the notion of an infinitesimal change in a function should itself be a function of all the possible infinitesimal changes of the point that the function is evaluated at, i.e. it should be a function of the all possible tangent vectors in the vector space tangent to that point (as such quantities describe the possible directions that a function can pass through a point and all the possible "speeds").  In this sense the differential change in the function as it passes through the point in a given direction should be equal to the directional derivative of the function along the particular vector describing that direction?!

REPLY [2 votes]: Like always, the intuition comes from the Euclidean case — here even the one-dimensional case!
In the differential geometry of the line $\mathbb{R}$ with coordinate function $t$, we have

The standard vector field, usually denoted $\partial /\partial t$
The standard differential form, denoted $\mathrm{d} t$

The standard vector field captures the geometric aspect of one-variable calculus and the standard differential form captures the algebraic aspect.
These applications may be a little hard to tell apart at first, but it becomes more clear when you see how the line is used:

Maps $\mathbb{R} \to M$ describe curves on a manifold. Each curve pushes the standard vector field $\partial / \partial t$ forward to give the tangent vectors to the curve on $M$.
Maps $M \to \mathbb{R}$ describe scalar fields on a manifold. Each scalar field pulls the standard differential form $\mathrm{d} t$ back to give the differential of the scalar field on $M$.

Now, we know how to combine differential forms and vector fields on $\mathbb{R}$; e.g. the simplest case is that if $y = g(t)$, then $\mathrm{d} y \cdot  \frac{\partial}{\partial t} = g'(t)$. This extends to the manifold: if we have a curve $\gamma$ and a scalar field $f$, then
$$ f^*(\mathrm{d} t) \cdot \gamma_*(\partial / \partial t) = (f \circ \gamma)'$$
This is consistent with the other two places we could do the combination:
$$ \mathrm{d} t \cdot (f_* \circ \gamma_*)(\partial / \partial t) = (f \circ \gamma)'$$
$$ (\gamma^* \circ f^*)(\mathrm{d} t) \cdot \partial/\partial t = (f \circ \gamma)'$$

As a side effect of my algebraic tendencies, I personally prefer a "differential forms first" approach to this sort of thing; IMO expressing things from the algebraic perspective tends to make calculation cleaner and more intuitive. In this approach, one might define the tangent vectors as linear functionals on the differentials.
e.g. $\mathrm{d}(x^2 y) = 2xy \mathrm{d}x + x^2 \mathrm{d} y$, and in $(x,y)$ coordinates, the vector $\partial / \partial x$ is the mapping that sends $\mathrm{d}x \to 1$ and $\mathrm{d} y \to 0$. I think this formulation is even a more direct description of what we mean by the notation $\partial / \partial x$: "the variation along with $x$ as $y$ is held constant".
I've never seen the foundations of differential geometry developed in this way, but I imagine you would start define the cotangent bundle in a manner analogous to however you would have defined the tangent bundle.
However, in other subjects where functions play more prominence at the foundational level, such as algebraic geometry, differential forms first is a standard approach.