Simplified and closer to your original question example

The First value of each sublist is compared and the sublist with the largest first value is returned

l1 = {{1, a}, {2, b}};
MaximalBy[l1, First]
(* {{2, b}} *}

In your question q is not a list of sublists, it has 3 levels of { }. 3 is > 1 so the second sublist is returned.

l2 = {{{1, a}, {2, b}}, {{3, c}, {4, d}}};
MaximalBy[l2, First]
(* {{{3, c}, {4, d}}} *}

There is no pairwise comparison between the elements in the sublists and only the First element of the sublist is considered. This does not change the result

l2 = {{{1, a}, {20, b}}, {{3, c}, {4, d}}}
MaximalBy[l2, First]
(* {{{3, c}, {4, d}}} *)

Regarding #, &, /@, read this and this.

It was not clear that you wanted to do a pairwise comparison between the two lists and pick the one with the largest first value. For that you need to combine them into pairs first and map.

q = Transpose[{y01, y11}];
MaximalBy[#, First] & /@ q

(* {{{4.9375, {x -> -0.75}}}, {{7., {x -> -2.}}}, {{15., {x -> -2.}}}} *)

You need to Join the two lists, not embed them in another list.

q = Join[y01, y11];
MaximalBy[q, First]

(* {{15., {x -> -2.}}} *)