Ты имеешь в виду это?
apply(dF[seq(1, ncol(dF), by = 2)], 1, function(x)
names(dF[seq(1, ncol(dF), by = 2)])[order(x)])[, 1]
#[1] "B1" "A1" "C1"
Или получить matrix
имен столбцов для всех 6 строк.
t(apply(dF[seq(1, ncol(dF), by = 2)], 1, function(x)
names(dF[seq(1, ncol(dF), by = 2)])[order(x)]))
# [,1] [,2] [,3]
#[1,] "B1" "A1" "C1"
#[2,] "A1" "B1" "C1"
#[3,] "A1" "C1" "B1"
#[4,] "C1" "A1" "B1"
#[5,] "C1" "B1" "A1"
#[6,] "C1" "B1" "A1"
Объяснение: dF[seq(1, ncol(dF), by = 2)]
выбирает столбцы 1, 3, 5, ... и apply(..., 1, function(x) names(dF[seq(1, ncol(dF), by = 2)])[order(x)])
возвращает имена столбцов упорядоченных записей по строкам.