A (too?) non-rigorous/intuitive approach:
Instead of dealing with all the children at once, consider there are two children 1 and 2, while all the other children are treated as 'external world', denoted by $E$. In each round $i$, $E$ gives 1 some coins, say $\alpha_i$. 1 gives half of its coins to 2, and two gives half of his to $E$ (after taking care of odd-numbered scenarios with the teacher's help). In order to model all possible behavior of the $n-2$ children, no restriction is imposed on $\alpha_i$s except they be natural numbers.
Suppose at any instant of time, say beginning of round 0, 1 and 2 had $(x_1,x_2)$ coins and it is 1's turn to share half his coins. Their coins change as:
Round 0: $(x_1,x_2) \to (\frac{x_1}{2},x_2+\frac{x_1}{2})\to (\frac{x_1}{2},\frac{x_2}{2}+\frac{x_1}{4})$
At the beginning of round 1, $x_1$ gets $\alpha_1$ coins from his neighbor in $E$. Effect of round 1 on children 1 and 2 is
Round 1:$(\frac{x_1}{2}+\alpha_1,\frac{x_2}{2}+\frac{x_1}{4})\to (\frac{x_1}{4}+\frac{\alpha_1}{2},\frac{x_1+x_2}{2} +\frac{\alpha_1}{2}) \to(\frac{x_1}{4}+\frac{\alpha_1}{2},\frac{x_1+x_2}{4} +\frac{\alpha_1}{4})$
Consider one more round, with 1 getting $\alpha_2$ coins this time.
Round2: $(\frac{x_1}{4}+\frac{\alpha_1}{2}+\alpha_2,\frac{x_1+x_2+\alpha_1}{4}) \to (\frac{x_1}{8}+\frac{\alpha_1}{4}+\frac{\alpha_2}{2},\frac{3x_1}{8} +\frac{x_2}{4}+\frac{\alpha_1+\alpha_2}{2}) \to (\frac{x_1}{8}+\frac{\alpha_1}{4}+\frac{\alpha_2}{2},\frac{3x_1}{16} +\frac{x_2}{8}+\frac{\alpha_1+\alpha_2}{4})$
All the coefficients for a particular term obey the series $f_i = \frac{1}{2}(f_{i-1}+\frac{1}{2^i}),f_0=0$ It is easy to see that $f_{i-1} and $f_i\to 0$ as $ i\to \infty$.
So, the amount of coins with 1 and 2 at the end round $i$ can be written as $x_1(i) = \frac{x_1}{2^i} +\sum_{j} \frac{\alpha_j}{2^{i-j}}$ $x_2(i) = \frac{x_2}{2^i} + f_ix_1 + \sum_{j}f_{i-j}\alpha_j$
It is easy to see that as $i$ increases, $x_1(i)-x_2(i)$ decreases. So, after some iterations, they will be close enough for the ceiling function to map to the same integer.
If this is granted, the rest of the result is easy to prove. By considering all pairings of adjacent children, it is easily shown that all of them will have the same number of coins once they have shared half of their coins. It follows that the child about to share his coins must have double the coins everybody has.
Note: This answer is just to present an intuitive understanding of what is happening after each round. I have neglected the ceiling function while analyzing the rounds, although ceilings and floors usually hasten convergence, by ironing out small differences. What remains to be proved is that any two adjacent children will eventually have the same number of coins, even when odd numbers are taken care of by ceiling/teacher. Working on that!!