Understanding `echo $((0x63))`

$(...) is a command substitution (not just a subshell), but $((...)) is an arithmetic expansion.

When you use $((...)), the ... will be interpreted as an arithmetic expression. This means, amongst other things, that a hexadecimal string will be interpreted as a number and converted to decimal. The whole expression will then be replaced by the numeric value that the expression evaluates to.

Like parameter expansion and command substitution, $((...)) should be quoted as to not be affected by the shell's word splitting and filename globbing.

echo "$(( 0x63 ))"

As a side note, variables occurring in an arithmetic expression do not need their $:

$ x=030; y=30; z=0x30
$ echo "$(( x + y +x ))"
78

This is not a subshell, but arithmetic evaluation. From man bash:

((expression))

The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to let "expression".