When you evaluate arithmetic expressions (like let does), a leading 0 indicates an octal number. You can force bash to use a given base by prefixing base#
to the number.
In addition, you can use printf to pad numbers with leading zeroes.
So your example could be rewritten as
cursorDay=2
let cursorDay=10#$cursorDay+1
printf -v cursorDay '%02d\n' "$cursorDay"
echo "$cursorDay"
or even shorter as
cursorDay=2
printf -v cursorDay '%02d\n' $((10#$cursorDay + 1))
echo "$cursorDay"
Please note that you cannot omit the $
between the #
and the variable name.