My take on this would be via the intuitive understanding of an Ito Integral. I feel it's best to interpret the Ito Integral via relating it to a gambling game: the integrator (i.e. the Brownian motion with respect to which we are integrating) is the (random) outcome of the gambling game, whilst the integrand (the function we are integrating) is the betting strategy. The betting strategy can be deterministic or random.
By design, at each point in time when the betting strategy is placed, the (random) outcome of the gambling game is not yet known, similarly to playing a roulette in a casino (hence why the integrator has to be forward-looking: by design, when the bet is placed (i.e. $f()$ becomes known), the game outcome (i.e. the integrator $W(t)$) is not yet known.
I believe that we can construct the Ito Integral both: (a) from the better's time point of view as well as (b) from the casino's time point of view:
(a) Ito Integral from the better's time point: let $f(\omega_{t_i},t_i)$ be the (possibly random) bet at time $t_i$, with $f(t_0)$ being the initial bet and $\omega_{t_i}$ denoting some random outcome at time $t_i$ ($\omega_t$ is adapted to the same filtration as $W_t$).
Indeed, the bet could be deterministic and even constant, in which case $f(\omega_{t_i},t_i)=k$, or it could be related to the outcomes that gradually become known as the game goes on, i.e. $f(\omega_{t_i},t_i)=f(W_{t_i})$
In general:
$$ I(f(\omega_h,h))_{h=t_0}^{h=t_n}:=\int_{h=t_0}^{t_n}f(\omega_h,h)dW(h)=\lim_{n \to\infty}\sum_{i=0}^{i=n-1}f(\omega_{t_i},t_i)\left(W(t_{i+1})-W(t_i) \right)$$
Above, at each time point, the better places a bet but does not yet know the random outcome of the game at the next time point.
(b) Ito Integral from the casino's time point: let $f(\omega_{t_i},t_i)$ be the (possibly random) bet at time $t_i$, with $f(t_0)$ being the initial bet. Then:
$$ I(f(\omega_h,h))_{h=t_0}^{h=t_n}=\lim_{n \to\infty} \sum_{i=1}^{i=n}f(\omega_{t_{i-1}},t_{i-1})\left(W(t_{i})-W(t_{i-1}) \right)$$
Above, at each time point, the casino knows the outcome of the random game, but it had known the better's bet before the random game had commenced.
Bottom line: intuitively, the expected value of the Ito integral is zero, because the integrator (i.e. the random game) is (by design) independent of the betting strategy. Since the integrator is a sum of independent Brownian motion increments, the expected value of Ito integral has to be zero, i.e.:
$$\mathbb{E}[I(f(\omega_h,h))_{h=t_0}^{h=t_n}]\approx \mathbb{E} \left[ \sum_{i=0}^{i=n-1}f(\omega_{t_i},t_i)\left(W(t_{i+1})-W(t_i) \right) \right]=\\=\sum_{i=1}^{i=n}\mathbb{E}[f(\omega_{t_i},t_{i-1})] \mathbb{E}[\left(W(t_{i})-W(t_{i-1}) \right)]=\\=\sum_{i=1}^{i=n}\mathbb{E}[f(\omega_{t_i},t_{i-1})] *0=0$$