$B_t$ is Brownian motion. It is assumed that motion starts at $0$.
I do not understand how the highlighted equalities hold true.
Is the first one equivalent to $E[(B_{t_{j}}-B_0)(B_{t_{j+1}}-B_{t_{j}})]$, where $B_0=0$? For the second one I tried expanding the expression to no avail. Not sure what $E[B_{t_{j}}^2]$ would equal and that does not seem to be the direction of thinking anyway.
For every $s\lt t$, $E(B_s(B_t-B_s))=E(B_s)E(B_t-B_s)=0$ because $B_s$ and $B_t-B_s$ are independent and $B_t-B_s$ is centered.
A consequence is that $E(B_t(B_t-B_s))=E(B_t(B_t-B_s))-E(B_s(B_t-B_s))=E((B_t-B_s)^2)$. Finally, $B_t-B_s$ is centered normal with variance $t-s$ hence $E((B_t-B_s)^2)=t-s$.
The hypothesis that one starts from $B_0=0$ is not necessary.
Prerequisite: The increments of Brownian motion are independent and centered normal with known variance.