Misunderstanding of proof with transitivity of logspace reduction
$L_f$, by definition, gives you the output of $f(x)$ for a single bit $j$. Note that we do not have to copy $x$, so we're only storing $bin(j)$ and the $j$-th bit of  $f(x)$, which is in logarithmic space. Since $M_f$ computes $L_f$ with logarithmic space, we can simply get the $j$-th bit. This means that $M_h$ can work in logspace in total, as it only asks and stores a single bit of input from $M_f$ every time.