Expectation of $dX_t$ for $X_t$ being an Ito process Let $X_t$ be an Ito Process:
$$dX_t = f(t, X_t)dt + g(t, X_t)dW_t$$
What is $E_t[dX_t]$?
How can we compute it and importantly what is the intuitive explanation of  $E_t[dX_t]$? 
 A: It should be pointed out that $E_t[dX_t]$ is a heuristic notation used by only non-mathematicians (e.g. empirical economists). It has no mathematical meaning. 
(Nor does statements like ``$E_t dX_t = 0$ if $X_t$ is a martingale".)
The correct mathematical formulation is via the infinitesmal generator:
$$
\lim_{\Delta t \rightarrow 0} \frac{E_{t+\Delta t} [ h(X_{t+\Delta t}) ] - E_t[h(X_t)]}{\Delta t}
= \mathcal{L}h(X_t).
$$ 
This describes the infinitesmal evolution of the conditional mean of $h(X_t)$ conditional on time $t$.
It is a standard fact, by Ito's lemma, that $\mathcal{L}h(X_t)$ is a second order differential operator
$$
\lim_{\Delta t \rightarrow 0} \frac{E_{t+\Delta t} [ h(X_{t+\Delta t}) ] - E_t[h(X_t)]}{\Delta t}
= h'(X_t) f(X_t) + \frac{1}{2} h''(X_t) g^2(X_t).
$$ 
If $h(x) = x$, this becomes 
$$
\mathcal{L}h(X_t) = f(X_t).
$$
Translated to  your heuristic notation,
$$
E[dX_t] = f(X_t) dt.
$$
A: $$
E_t[dX_t] = E[f(t, X_t)] dt + E_t[g(t, X_t)]E_t[dW_t] = E[f(t, X_t)] dt + E_t[g(t, X_t)] \times 0.
$$
You can interpret $E[f(t, X_t)]$ as how much the process moves, on averate, between $t$ and $t + dt$. Put more simply
$$
E_t[X_{t + dt}] = X_t + E_t[f(t, X_t)] dt.
$$
