I would suggest a tidyverse approach where there is no need of a function. In this solution you can extract the first letter of each variable name as id and then assign a number with cur_group_id so that the order is kept. Finally, with this new number you transform the variable containing the names and then you format to wide in order to obtain the expected output:

library(tidyverse)
#Data
df<-data.frame("id"=c(1,2,3,4),
               "ex"=c(1,0,0,1),
               "aQL"=c(5,4,NA,6),
               "bQL"=c(5,7,NA,9),
               "cQL"=c(5,7,NA,9),
               "bST"=c(3,7,8,9),
               "cST"=c(8,7,5,3),
               "aXY"=c(1,9,4,4),
               "cXY"=c(5,3,1,4))
#Reshape
df %>% pivot_longer(cols = -c(1,2)) %>%
  #Extract first letter as id
  mutate(id2=substring(name,1,1)) %>%
  #Create the number id
  group_by(id2) %>%
  mutate(id3=cur_group_id()) %>%
  #Clean name
  mutate(name=substring(name,2,nchar(name))) %>%
  #Create final var
  mutate(name2=paste0(name,'.',id3)) %>% ungroup() %>%
  dplyr::select(-c(name,id2,id3)) %>%
  #Format to wide
  pivot_wider(names_from = name2,values_from=value)

Output:

# A tibble: 4 x 9
     id    ex  QL.1  QL.2  QL.3  ST.2  ST.3  XY.1  XY.3
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     1     5     5     5     3     8     1     5
2     2     0     4     7     7     7     7     9     3
3     3     0    NA    NA    NA     8     5     4     1
4     4     1     6     9     9     9     3     4     4

in base R you could do:

names(df) <- sub("(\\d)([A-Z]{2})$","\\2.\\1", chartr("abc","123",names(df)))
 df
  id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
1  1  1    5    5    5    3    8    1    5
2  2  0    4    7    7    7    7    9    3
3  3  0   NA   NA   NA    8    5    4    1
4  4  1    6    9    9    9    3    4    4

If you need the NA columns:

names(df) <- sub("(\\d)([A-Z]{2})$","\\2.\\1", chartr("abc","123",names(df)))
a <- read.table(text=grep("\\.\\d",names(df),value = TRUE), sep=".")
b <- subset(aggregate(.~V1, a, function(x) setdiff(1:3,x)), V2>0)
df[do.call(paste, c(sep = ".", b))] <- NA
(df1 <- df[c(1, 2, order(names(df)[-(1:2)]) + 2)])

  id ex QL.1 QL.2 QL.3 ST.1 ST.2 ST.3 XY.1 XY.2 XY.3
1  1  1    5    5    5   NA    3    8    1   NA    5
2  2  0    4    7    7   NA    7    7    9   NA    3
3  3  0   NA   NA   NA   NA    8    5    4   NA    1
4  4  1    6    9    9   NA    9    3    4   NA    4

Here is a solution that uses regular expressions via the stringr package:

library(stringr)

df<-data.frame("id"=c(1,2,3,4),
               "ex"=c(1,0,0,1),
               "aQL"=c(5,4,NA,6),
               "bQL"=c(5,7,NA,9),
               "cQL"=c(5,7,NA,9),
               "bST"=c(3,7,8,9),
               "cST"=c(8,7,5,3),
               "aXY"=c(1,9,4,4),
               "cXY"=c(5,3,1,4))

renameCol <- function(x) {
  col_names <- colnames(x)
  index_ql <- str_detect(col_names,
                         "^[a-z]{1}QL")
  index_st <- str_detect(col_names,
                         "^[a-z]{1}ST")
  index_xy <- str_detect(col_names,
                         "^[a-z]{1}XY")
  
  replace_fun <- function(x) {which(letters %in% x)}
  
  col_names[index_ql] <- paste0("QL.", str_replace(substr(col_names[index_ql], 1, 1),
                                                  "[a-z]", replace_fun))
  col_names[index_st] <- paste0("ST.", str_replace(substr(col_names[index_st], 1, 1),
                                                   "[a-z]", replace_fun))
  col_names[index_xy] <- paste0("XY.", str_replace(substr(col_names[index_xy], 1, 1),
                                                   "[a-z]", replace_fun))
  
  col_names
  
}

colnames(df) <- renameCol(df)

df
#>   id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
#> 1  1  1    5    5    5    3    8    1    5
#> 2  2  0    4    7    7    7    7    9    3
#> 3  3  0   NA   NA   NA    8    5    4    1
#> 4  4  1    6    9    9    9    3    4    4

^{Created on 2020-09-07 by the reprex package (v0.3.0)}

Edit

The function above was adapted so that it takes the order into account.

Another way you can try

colnames(df)[grepl("QL", colnames(df))] <- str_c("QL.", 1:3)

colnames(df)[grepl("ST", colnames(df))] <- str_c("ST.", 2:3)

colnames(df)[grepl("XY", colnames(df))] <- str_c("XY.", c(1,3))

#   id ex QL.1 QL.2 QL.3 ST.2 ST.3 XY.1 XY.3
# 1  1  1    5    5    5    3    8    1    5
# 2  2  0    4    7    7    7    7    9    3
# 3  3  0   NA   NA   NA    8    5    4    1
# 4  4  1    6    9    9    9    3    4    4

using base pattern matching:

you need to define a function that does what you want on one single column name:

f = function(x){
  beg <- str_extract(x,"[a-z](?=[A-Z]{2})")
  num <- which(letters == beg)
  output <- paste0(str_extract(x,"(?<=[a-z])[A-Z]{2}"),".",num)
  return(output)
}

here extract the lower case letter if you have two upper case letters after, find it position in alphabet, and paste the found number back to the upper case letters.

> f("cQL")
[1] "QL.3"

You can then use regmatches and regular expression directly on the name of your data frame:

m <- gregexpr("[a-z][A-Z]{2}", names(df),perl = T)
regmatches(names(df), m) <- lapply(regmatches(names(df), m), f)
names(df)

> names(df)
[1] "id"   "ex"   "QL.1" "QL.2" "QL.3" "ST.2" "ST.3" "XY.1" "XY.3"

It solves only the renaming part, not the the "including missing column number" part of your question