As you suspect, you have a mistake. You say that:

$$R_{average} = W^{1/n} = (\prod_{i = 1..n}R_{i})^{1/n} = {1\over n} \sum_{i = 1..n}log(R_{i}).$$

Notice that you took a log and kept the equation sign. What you really meant is

$$\log R_{average} = {1\over n} \sum_{i = 1..n}log(R_{i}).$$ So you don't really compare

$$\max_{f} W \space \space \space \space \space vs. \space \space \space \max_{f} R_{average},$$

but rather $$\max_{f} W \space \space \space \space \space vs. \space \space \space \max_{f} \log R_{average}= \frac 1 n \max_{f} \log W.$$

Finally, what confuses you is that

But, it should be the same thing, right? The parameters (the betting fraction) to optimise $W^{1/n}$ should be the same as to optimise $W$.

You see, $\text{argmax}_f W \neq \text{argmax}_f \log W $ in the general case (and also in Kelly's).

As you suspect, you have a mistake. You say that:

$$R_{average} = W^{1/n} = (\prod_{i = 1..n}R_{i})^{1/n} = {1\over n} \sum_{i = 1..n}log(R_{i}).$$

Notice that you took a log and kept the equation sign. What you really meant is

$$\log R_{average} = {1\over n} \sum_{i = 1..n}log(R_{i}).$$ So you don't really compare

$$\max_{f} W \space \space \space \space \space vs. \space \space \space \max_{f} R_{average},$$

but rather $$\max_{f} W \space \space \space \space \space vs. \space \space \space \max_{f} \log R_{average}= \frac 1 n \max_{f} \log W.$$

Finally, what confuses you is that

But, it should be the same thing, right? The parameters (the betting fraction) to optimise $W^{1/n}$ should be the same as to optimise $W$.

Remember that when you optimize, you care about $\mathbb E[W]$. You see, $\text{argmax}_f \mathbb{E}[W] \neq \text{argmax}_f \mathbb{E}[\log W] $.