Power of Double - Logarithmic Space
Well, the set $\left \{w_1**w_2**..**w_k \right \}$ contains only one single element, but is nevertheless a set. And TM can accept only languages, which are sets of words.

Why is it slightly easier? Hmm, I would say, it is easier since you can simplify the algorithm slightly. You have one counter for $k$, and then you check, whether in the string $\cdots *x*y*\cdots$, you have $\text{bin}(y)=\text{bin}(x)+1$ on every position. You do not have to care about detecting the right $k$. Other than this, the suoer-string is longer, which gives you more space. But this effect vanishes in the big-O.