Least square regression test statistic

For a regression

$$y_t = x_t' \beta + u_t,$$

let $H$ be the hat matrix such that $\hat y = Hy$.

Let $M = I - H$.

I am trying to understand the proof $E[\hat u (b-\beta)'] = 0$.

I was able to understand $E[\hat u (b-\beta)'] = E[Muu'X(X'X)^{-1}]$, but couldnt understand why the following step holds $$E[Muu'X(X'X)^{-1}] = \sigma^2MX(X'X)^{-1} = 0$$ Why is this the case? Could anyone please help?

For the first equality, note that - this is not stated explicitly in your assumptions, but likely mentioned somewhere in your source - $E(uu')=Cov(u)=\sigma^2I$. Also, it appears that $X$ is treated as nonstochastic, so that it may be taken out of the expectation operator. Hence,
$$E[Muu'X(X'X)^{-1}] = ME[uu']X(X'X)^{-1} = M\sigma^2IX(X'X)^{-1}= M\sigma^2X(X'X)^{-1}$$
For the second equality, recall that $H=X(X'X)^{-1}X'$, so that $$MX=(I-H)X=X-X(X'X)^{-1}X'X=X-X=0.$$
From an intuitive perspective, $M$ is the "residual-maker matrix". That is, for a vector $y$, $My$ gives the residuals of regressing $y$ on $X$. Hence, the columns of $MX$ give the residuals of regressing each column in $X$ on $X$. That is, we consider the residuals ("mistakes made") when explaining a column of $X$ by, among amothers, itself. Now, it is intuitive that you do not make a mistake (get zero residuals) when explaining something by itself.