Why can we neglect a product of differentials? In a physics problem, if we wish to find the differential change in the momentum of a rocket losing mass but gaining velocity, we may write that
$$\textrm{d}p=(m-\textrm{d}m)(v+\textrm{d}v)-mv.$$
However, upon expansion, we choose to neglect the "product of differentials" $\textrm{d}m\textrm{ d}v$ and write
$$\textrm{d}p=m\textrm{ d}v-v\textrm{ d}m.$$
Why are we allowed to do this? If this is allowed, wouldn't any double integral, $\iint_{R}{f(x,y)\textrm{ d}A}$, vanish since in some sense, we define $\textrm{d}A=\textrm{d}x\textrm{ d}y$?
In essence, when do I neglect differentials and is there a rigorous discipline for the treatment of differentials?
 A: There are several possible answers, depending on what your precise question is.


*

*The differential $\text d$ is a first-order, well, differential operator. Hence, if $p=mv$, then $\text d p=\text{d}\left(mv\right)=\left(\text{d}m \right)v + v \,\text{d}m$.

*If you think of $\text{d}x$ as a "very small" change in $x$ (basically like  $\Delta x$ but smaller), then $\text{d}m \text{d}v$, being second order in the small quantities, will be much smaller than the first-order terms $m \,\text{d}v$ and $v \,\text{d}m$ and can be neglected to good approximation. Note that $m v$ is again much larger than each of these two, but the zero-order term gets subtracted. Thus the first-order terms are the leading ones.

*If you think about derivatives, you can make the previous step rigorous by expressing $\text{d}v$ and $\text{d}m$ in terms of, say, $\text{d}t$, dividing by $\text{d}t$ and taking a limit. Then the higher-order terms exactly vanish.

*This is not explicitly in your question, but note that the $\text{d}x\,\text{d}y$ etc. in an integral are (secretly) differential forms, which have a special antisymmetric multiplication -- that's why you can't have $\text{d}x\, \text{d}x$ integrals, and why you get determinants on changing the integration variables.
A: There's a fair bit of notational trickery going on here that hides the formal meaning of the notation - neither $(m-dm)(v+dv)$ nor $dx\,dy$ are actually products and, even worse, they're not even the same operation. Both, however, exist in an ecosystem of math in which we want to describe functions and their rates of changes along curves - but nothing more.
To understand the first thing, it's worth just remembering what momentum is defined as:
$$p=mv.$$
These may all be thought of as variables, and the usual product rule can be phrased as the equation
$$dp=m\,dv + v\,dm$$
which translates most literally to say that, along any path obeying the law $p=mv$, the rate of change in momentum is proportional to the current mass times the change in velocity plus the current velocity times the rate of change of mass (which is presumably negative). The product of the two rates of change has nothing to do with the rate of change of $p$, so it doesn't get mentioned.
Formally, what we should think about is that, if we considered a three dimensional space with coordinates $(p,m,v)$ and looked at the surface defined by the equation $p=mv$ - which is a sort of saddle shape (a hyperbolic paraboloid, to be technical) - we could imagine defining three functions $p,m,v$ taking points on this surface to the three coordinates. At any point on this surface, we could define a tangent plane, which represents, more or less, the possible velocities a point moving across the surface could have. 
Then, we do the calculus thing: if each point in this tangent plane represents a velocity, then we could calculate how fast the functions $p,\,m$ and $v$ change at a given point with a given velocity - and these functions being differentiable means that these these changes are linear in the velocity. We call the function that takes a point in these tangent planes to the change in the respective variable $dp,\,dm$ and $dv$. In general, these objects which assign each point on a surface to a linear function on the tangent space are called differential 1-forms - and you can integrate them along curves by the usual methods, summing up the evaluations of these linear functions on the instantaneous velocities of the curve. The equation
$$dp = m\,dv+v\,dm$$
simply means that, at every point, the linear function $dp$ is a certain combination of the linear functions $dv$ and $dm$ - and, given that the tangent space is two dimensional and we have three linear functions, there always will have to be some non-trivial relation among them. Note that the multiplication $p=mv$ is a real multiplication, but happens before anything differential happens.
Your notation is not quite compatible with this view, but it's informed by this theory: differential forms are first order approximations meaning that they only take into account linear terms. Even though $(1+x)(1-x)=1-x^2$, the best linear approximation to $1-x^2$ is still just $1$. That's why, if we say that, at a point, $m$ is approximated by $m+dm\cdot \Delta$ - imagined as a linear function -and $v$ is approximated by $v+\,dv\cdot \Delta$, the best linear approximation to $p$ is still $mv + (v dm + m\,dv)\cdot Delta$ even if, on the face of it, we might desire some quadratic term. This isn't just a definitional problem: we never presumed we knew anything about how $m$ or $v$ behaves beyond their linear terms, so we cannot extrapolate anything about the quadratic term of $mv$. For instance, if we approximate $1+x^2$ linearly by $1$ and approximate $1+x$ linearly by $1+x$, the product of the functions is $1+x+x^2+x^3$, but the product of the linearizations is $1+x$, so has gotten the quadratic term wrong, hence why we should discard that term.
Okay, but what's the $dx\,dy$ in integration? Well, that's a wedge product in the theory of differential forms - perhaps more properly written $dx\wedge dy$ in areas where we plan to use differential forms as more than just nice notation. To understand them, it's worth rephrasing what a differential form does:

A differential 1-form assigns a rate of change to infinitesimal pieces of a smooth curve.

Where "an infinitesimal piece of a smooth curve" is essentially interpreted as "a point and a velocity." For instance $dp$ assigns little pieces of a curve to the amount that the momentum changed over that little piece. An integral takes all of these little pieces and sums them up. So if we want to integrate over area, it's clear what we must do:

A differential 2-form assigns a rate of change to infinitesimal pieces of a parameterized surface.

It's a bit more difficult to formally describe what has to happen here; basically, one imagines that, in the tangent plane, a little piece of area looks like a (signed) parallelogram, and a differential 2-form is some sort of "linear" rule for assigning weights to these parallelograms. Informally, we want to consider the space of infinitesimal areas near a given point. Formally, one typically uses the exterior algebra (essentially the algebra of areas, volumes, and so on) to talk about this - but, for instance, a differential form like $dx\wedge dy$ says "project the parallelogram to the $xy$-plane and measure its area there." It doesn't say "multiply the change in $x$ with the change in $y$" because that doesn't make sense when we're working with surfaces rather than curves*. As before if we have a differential 2-form and a surface, we can integrate by summing up all the little bits, according to the rule specified by the differential 2-form.
The point of this whole theory of differential forms is that the entire theory is about making complicated functions into linear functions. It turns out, as a theorem, that when you do this, when rates of change meet each other in multiplication, they should be ignored because they don't affect the linearization of the product, and they can't be interpreted to mean anything about the product. This is the common formalization of differentials - and, though you can come up with theories that track more than linear data, that's not what the differential notation is used for.

*These differential forms act a bit differently than one might expect - the notation
$$\iint f(x,y) \,dx \,dy$$
is sometimes interpreted as two integrals as
$$\int \left(\int f(x,y)\,dx\right)\,dy$$
but this isn't really compatible with the differential form view of the world, since $f(x,y)\,dx\wedge dy$ is then a term in itself - and, as it turns out, $dy\wedge dx$ is "reflected" hence is the negative of $-dx\wedge dy$, which is a bit annoying if you had wanted to easily transition from notation in which order doesn't matter to the full theory of differential forms.
A: This is not an answer, just too long for a comment:
Let $p :\mathbb{R}^2 \to \mathbb{R}$ be given by $p(x) = x_1 x_2$, then we can compute the derivative using the following estimate
$p(x+h)-p(x) = (x_1+h_1)(x_2+h_2)-x_1x_2 = x_2 h_1 + x_1 h_2 + h_1 h_2$.
(Think of the $h_k$ as the 'differential terms'.)
We see that 
$|p(x+h)-p(x)-(x_2 h_1 + x_1 h_2)| \le |h_1 h_2| \le \|h\|^2$ from which we see that
the derivative is $Dp(x)h = x_2 h_1 + x_1 h_2$.
