Given E(u|x)=0, how do we get $E(E(xu|x))=E(xE(u|x))$?
I think as $x$ is conditional on a specific value, we get it reduced to $E(au|x=a)$, where $x$ and $u$ are variables but since it is given that $x$ can only be $a$, it acts like a constant and we can filter it out of the expectation operator. We use the rule: $E(aX)=aE(X)$, where $a$ is a constant and $X$ is a variable. Hence we only have to figure out $E(u|x)$ where $x$ is conditioned as equal to a certain value in this case "$a$".
I think intuitively this works but can someone please show me using the definition of the expectation operator i.e. using summation? As to how we can write $E(xu|x)=xE(u|x)$.
When conditioning random variable $Z$ over random variable $W$, then we may use the substitution: $$\mathsf E(Z\mid W)={\big[\mathsf E(Z\mid W{=}w)\big]}_{w=W}$$
When conditioning a random variable $Z$ over an event $\mathcal A$, then we use the normalisation: $$\mathsf E(Z\mid\mathcal A)=\mathsf E(Z\,\mathbf 1_\mathcal A)\,\mathsf P(\mathcal A)^{-1}$$
So putting this together we derive:
$$\begin{align}\mathsf E(XU\mid X)&={\big[\mathsf E(XU\mid X=x)\big]}_{x=X}\\[1ex] &={\big[\mathsf E(\mathbf 1_{X=x}XU)\,\mathsf P(X=x)^{-1}\big]}_{x=X}\\[1ex]&={\big[\mathsf E(x\,\mathbf 1_{X=x}\,U)\,\mathsf P(X=x)^{-1}\big]}_{x=X}\\[1ex]&={\big[x\,\mathsf E(\mathbf 1_{X=x}\,U)\,\mathsf P(X=x)^{-1}\big]}_{x=X}\\[1ex]&={\big[x\,\mathsf E(U\mid X=x)\big]}_{x=X}\\[1ex]&= X\,\mathsf E(U\mid X)\end{align}$$
Which is precisely your intuition expressed in symbolic form.
NB: $\mathbf 1_\mathcal A$ is the indicator of the event, equalling $1$ when the event occurs, and $0$ otherwise. $$\mathbf 1_\mathcal A=\begin{cases}1&:&\mathcal A\\0&:&\text{otherwise}\end{cases}$$