How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$? That's how it's framed in my Physics school-book.
The question (or rather, the explanation) is that of the thrust of rockets and how the impulse is equal (with opposite signs) on the thrust-gases and the rocket itself.
$( m \frac { \Delta v}{ \Delta t} ) = -(v \frac {\Delta m}{ \Delta t} ) = F_i$ 
I suppose it's a problem with how I see the transfer of impulse and exactly which part of the equation relates to which part of the physical world (gases, rocket). So we can start from there.
Title equation:
$F = \frac{ \Delta (mv)}{ \Delta t} = \left ( m \frac { \Delta v}{ \Delta t} \right) + \left ( v \frac { \Delta m}{ \Delta t} \right)$
Grade: The equivalent of G-10 in the US.
 A: It's only true when the changes $\Delta t$, $\Delta v$, $\Delta m$ are small and then it is known as the Leibniz rule, the rule for the derivative of a product, which Leibniz (but also Newton) discovered when they invented the calculus 3 centuries ago. 
Just look at this proof:
$$\frac{\Delta (mv)}{\Delta t} = \frac{(mv)_{new} - (mv)_{old}}{\Delta t} 
=\frac{(m_{old}+\Delta m)(v_{old}+\Delta v)-m_{old}v_{old}}{\Delta t} = \dots
$$
Here I just used $x_{new}=x_{old}+\Delta x$ which holds for $X=m,v,t,mv$ or anything else. The new value is the old value plus the increment.
But now, expand the parentheses' products via the distribution law. You get;
$$=\dots \frac{m_{old}\Delta v+ \Delta m v_{old}+\Delta m\,\Delta v}{\Delta t} $$
because the term $m_{old}v_{old}$ canceled (it was subtracted). Now, if each $\Delta X$ is significantly smaller than $X$, e.g. 100 times, then $\Delta m \,\Delta v$ is 10,000 times smaller and can be totally neglected. So you're only left with the first two terms in the numerator and they give you exactly the two terms you wanted to find.
So the increase of $mv$ is obtained either from an increase of $m$ or from an increase of $v$. The equation just encodes this simple observation quantitatively.
A: The idea about this school book derivation is that you can change moment by changing velocity (common case) or by changing mass.
Since you can change the momentum of the system (rocket plus gasses) only by the external force, and in case of the rocket there is no external force (neglect gravity for a moment), so the question is, why is rocket getting faster and faster?
$$F_\text{ext} = \frac{\text{d}p}{\text{d}t} = m \frac{\text{d}v}{\text{d}t} + v \frac{\text{d}m}{\text{d}t} = 0,$$
$$m \frac{\text{d}v}{\text{d}t} = - v \frac{\text{d}m}{\text{d}t}.$$
Important contribution comes from the fact that as rocket pushes gasses away it effectively decreases its mass (right side of the equation).  If momentum is conserved, this means that velocity of the rocket increases (left side of equation).
A: Here is a visualization:
Momentum is mass times velocity, so draw it as the area of a rectangle:

If we change the mass and velocity a little, we change the momentum:

The total change in the momentum is the sum of green, blue, and purple rectangles. Their sizes are just length times width, so overall we have
$\Delta p = m\Delta v + v\Delta m + \Delta v \Delta m$
This looks like the answer you're seeking except for the extra term at the end.
Suppose we cut $\Delta m$ and $\Delta v$ down to one tenth their current size. Then the first two terms become one tenth as large, but $\Delta m \Delta v$ becomes one hundredth as large. The purple box shrinks away much faster than the blue and green ones. Therefore, for very small changes, we can ignore the purple box and write
$\Delta p = \Delta(mv) = m\Delta v + v\Delta m$
we usually indicate this limiting procedure by changing the $\Delta$ to $\mathrm{d}$, so
$\mathrm{d} p = \mathrm{d}(mv) = m\mathrm{d} v + v\mathrm{d} m$
