Algebraic orders of sum and product 
Suppose $f(z)$ and $g(z)$ have algebraic orders $n$ and $m$ at $z=0$, respectively. Estimate the algebraic orders of the product $f(z)g(z)$ and the sum $f(z)+g(z)$, and give examples to show that the bounds are sharp.

The algebraic orders mean that $f(z)=z^nf_1(z)$ and $g(z)=z^mg_1(z)$ for all $0<|z|<R$, where $f_1(z)$ and $g_1(z)$ are nonzero and holomorphic in $|z|<R$. 
I think the order of the product should be $n+m$, since $f(z)g(z)=z^{n+m}f_1(z)g_1(z)$, and $f_1(z)g_1(z)$ is nonzero and holomorphic in $|z|<R$.
For the sum, suppose without loss of generality that $n\geq m$. Then $f(z)+g(z)=z^m(z^{n-m}f_1(z)+g_1(z))$. So the algebraic order is at least $m$. If $n>m$ then $$z^{n-m}f_1(z)+g_1(z)\neq 0$$ so the order is exactly $n$. Now if $n=m$ then $f(z)+g(z)=z^n(f_1(z)+g_1(z))$. What can we say about the order of $f_1(z)+g_1(z)$?
 A: As you've shown, the order of $f(z)g(z)$ at $z=0$ is the sum of the orders of $f(z)$ and $g(z)$.
As for the product, the following is true: $\operatorname{ord}_{z=0} f(z)g(z) \geq \operatorname{min}\{\operatorname{ord}_{z=0} f(z), \operatorname{ord}_{z=0} g(z)\}$.
You pretty much wrote the proof yourself.  If $f(z) = z^n f'(z)$, and $g(z) = z^n g'(z)$, then $f(z)+g(z) = x^n (f'(z)+g'(z))$, which means that if both orders are at least $n$, then the sum is at least $n$.  In other words, the order of the sum must be at least the minimum of the two orders.
When the orders are not equal, we can say a little more—and you have.  If $n>m$, $f(z) = z^n f'(z)$, and $g(z) = z^m g'(z)$, where $f'(0)\neq 0$ and $g'(0) \neq 0$, then $f(z)+g(z) = x^m (f'(z) + x^{n-m}g'(z))$.  We can check that the inner polynomial is nonzero at $z=0$, so $\operatorname{ord}_{z=0} f(z)g(z) = \operatorname{min}\{\operatorname{ord}_{z=0} f(z), \operatorname{ord}_{z=0} g(z)\}$ in this case.
Another way of saying this is to say that, of the three numbers $\operatorname{ord}_{z=0} f(z), \operatorname{ord}_{z=0} g(z), \operatorname{ord}_{z=0} fg(z)$, it must be that the smallest number appears twice (of course, all three can be equal).
This should completely answer the question, but, since there's a bounty, here's a bit of background.  All of the above is saying that the ring of functions "near $0$" is a discrete valuation ring.  A "valuation", in this context, is something that satisfies the equations $\nu(ab)=\nu(a)\nu(b)$ and $\nu(a+b) \geq \operatorname{min}\{\nu(a),\nu(b)\}$.  $\nu(0)$ is given the special value $+\infty$ to make it all work.
Interestingly enough, you can use a valuation to make the set of complex analytic functions into an interesting (pseudo)metric space: $|f| = e^{-\operatorname{ord}_{z=0} f(z)}$.  This is, in fact, an ultrametric space, satisfying the strong triangle inequality: $|f+g| \leq \operatorname{max}\{|f|,|g|\}$, which means that every triangle is isoceles!
(Yes, I'm working with norms, not metrics.  So sue me.)
