Consider Brownian motion $W=\{W(t)\}$ with its natural filtration $\{F(t)\}$, $t \in [0,T]$ Furthermore consider the process $X=\{X(t)\}$ is $F(t_{k-1})$ measurable with $0=t_0<t_1<.....<t_n=T$, where $X(t)=X_k$ for $t \in [t_{k-1},t_k]$ and $k=1,..,n$. Why can I state the following:
$$\mathbb{E}\left[X_k \left(W(t_k)-W(t_{k-1})\right)^2\right]= \mathbb{E}\left[X_k \mathbb{E}\left[(W(t_k)-W(t_{k-1}))^2| F(t_{k-1})\right]\right] $$
What theorem do I use to insert the extra cond. expectation?
It is a theorem known as the law of total expectation that, for a probability space $(\Omega,\mathcal{F},\mathbb{P})$, for any sub-sigma algebras $\mathcal{G_1} \subseteq \mathcal{G_2} \subseteq \mathcal{F}$, you have $$ \mathbb {E} [\mathbb{E} [X\mid {\mathcal {G}}_{2}]\mid {\mathcal {G}}_{1}]=\mathbb{E} [X\mid {\mathcal {G}}_{1}]\quad {\text{(a.s.)}}$$ (the proof is in the linked article)
In your case, you just have to apply it for $\mathcal{G_1} =\sigma(\Omega) = \{\varnothing, \Omega \} $ and $\mathcal{G_2} = F(t_{k-1})$ to get your result.