Judging from the captions of the pictures, I think we should still talk about real derivatives for a bit.
Brief answer
Neither of functions depicted in your graphs are going to be differentiable at the discontinuities depicted. After you fill in a removable discontinuity of a function like the one on the left, it could be either differentiable or nondifferentiable at the point. Jump discontinuities of functions on the real line are always nondifferentiable, but they might have one-sided derivatives that are well-defined.
Longer anwer
First of all, remember that the derivative at a point is, intuitively, a "limit of slopes as calculated from the left and from the right." From the left you take a limit of $\frac{f(x)-f(x-h)}{h}$ over very small positive values of $h$, and on the right the same happens with $\frac{f(x+h)-f(x)}{h}$. (It can be the case that both of these can be defined, but they don't match and in that case, the derivative isn't defined at that point.)
Notice also that it is critical for $f(x)$ to be defined to carry out these computations, and so you won't get anywhere at all without settling on a value for $f(x)$. If you insist that there's no value for $f(x)$, then the slope is formally undefined. If you are willing to fill in removable discontinuities, though, you can proceed. The derivative may or may not exist after the point is filled in (consider $f(x)=|x|$ with the $x=0$ point removed/replaced.)
That leaves the case of the jump discontinuity, which you've depicted in the right hand picture. Jump discontinuities always make one of the slope limits on the right or on the left jump to infinity. Here's what I mean. Suppose $f(x)$ is anywhere exept filling in the lower circle in your right hand picture. Then as you shrink $h$ in $\frac{f(x+h)-f(x)}{h}$, the associated picture is that of a line which always lies on $(x,f(x))$ and $(x+h,f(x+h)$, which lies on the branch on the right. You see as $h$ shrinks, $x+h$ approaches $x$ from the right. Since $f(x)$ is not on that lower empty circle, this line tips ever more steeply as $h$ shrinks. Thus its slope goes to either $+\infty$ or $-\infty$, and the slope there is undefined.
If $f(x)$ happened to land on that empty lower right circle, then you are guaranteed it wouldn't land on the upper left circle, so you would then deduce that the slope estimate from the left would go off to infinity, and the derivative at the point would again not exist.
The claim$^1$ is false: the operation you've described doesn't always output a Eudoxus magnitude. In particular - and shifting from functions to sequences for ease of writing - take $$f=(1,2,5,10,21,...),\quad g=(1,3,6,13,26,...)$$ (the important feature being that each doubles infinitely often, but they never double at the same time). Then when we try to build their "Eudoxus sum" according to your rule, we have good news and bad news. On the plus side, the "initial segment sums" we get agree with each other; however, on the minus side we get the sequence $$(2,5,11,23,37,...)$$ which never doubles and hence fails to satisfy condition $(2)$.
Dropping condition $(2)$ from the definition doesn't help: the sequence $$(1,3,7,15,31,...)$$ would then be considered a Eudoxus magnitude, but there's no good way to define "Eudoxus sum" so as to not have its Eudoxus sum with itself violate $(1)$.
What you need is a way to ensure frequent simultaneous doubling. You can ensure this by strengthening condition $(2)$; e.g. you'll get better results if you demand that $f$ is a Eudoxus magnitude only if its set of doubling points has asymptotic density $1$ (since the intersection of two asymptotic-density-$1$ sets has asymptotic density $1$). But I suspect such a requirement will ruin things elsewhere, namely the original motivation for these.
$^1$This is a bit unfair of me. Your actual claim as written is just that we never see disagreement between the "initial segment sums," and it's not hard to show that this is in fact true. But implicitly, you want the output to again be a Eudoxus magnitude ("A bonus is that we can then immediately assert that the [Eudoxus sum] of two Eudoxus magnitudes is a Eudoxus magnitude.") and this doesn't hold.
Best Answer
A binary operation, namely $* : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}, (x,y) \mapsto x*y$ in the present case, is nothing else than a bivariate map $f(x,y) := x*y$. Since $x$ and $y$ are here themselves functions of another variable, let's say $t$, the chain rule permits to write : $\frac{\mathrm{d}}{\mathrm{d}t} (x(t)*y(t)) = f_x\dot{x} + f_y\dot{y}$, where subscripts denote partial derivatives.