For a fixed $d$, you just repeat the process of differentiation $d$ times, recursively. For example, if you want to get to $d = 3$, start with the first difference:
$$\Delta X_t = X_t - X_{t-1} =: Y_t$$
Then, $Y_t$ is just some series, which you can difference just like you did $X_t$. That would be the 2nd difference:
$$\Delta^2 X_t = \Delta (\Delta X_t) = \Delta Y_t = Y_t - Y_{t-1} = (X_t - X_{t-1}) - (X_{t-1} - X_{t-2}) = X_t - 2X_{t-1} + X_{t-2}$$
And then once more to get to the third difference:
$$ \Delta^3 X_t = \Delta (\Delta^2 X_t) = (X_t - 2X_{t-1} + X_{t-2}) - (X_{t-1} - 2X_{t-2} + X_{t-3}) = X_t - 3X_{t-1} + 3 X_{t-2} - X_{t-3}$$
So, $\Delta^dX_t$ is indeed a linear combination of $X_t, X_{t-1}, ..., X_{t-d}$, but not the one you suggested.
In an ARIMA(p,d,q) model, $\Delta^dX_t$ follows an ARMA(p,q) process, that's correct.