I think that this is what I'd do, although it returns the answer in a slightly different form than you've asked for - my approach is to avoid ragged arrays (ones with different column lengths).
Start with your data:
d <- c("GOEK, WOWP, PEOL, WJRN, KENC, QPOE, JFPG, PWKR, PWEOR, JFOKE, POQK, LSPF, PEKF,PFOW, VCNS, ALAO, LFOD",
"KFDL, LFOD, WOWP, PWEO, PWEOR, PRCP, ALPQ, JFOKE, ALLF, VCNS CNIR",
"KJTJ, FKOF, VCNS, FLEP", "FKKF, EPTR", "QPOE, PEOL, WJRN, VCNS, PEKF, PFPW" )
key <- structure(list(Items = c("ALAO", "ALLF", "ALPQ", "CNIR", "EPTR",
"FKKF", "FKOF", "FLEP", "GOEK", "JFOKE", "JFPG", "KENC", "KFDL",
"KJTJ", "LFOD", "LSPF", "PEKF", "PEOL", "PFOW", "PFPW", "POQK",
"PRCP", "PWEO", "PWEOR", "PWKR", "QPOE", "VCNS", "WJRN", "WOWP"
), A = c(NA, 0.016262491, 0.409770566, NA, 0.718354484, 0.801489091,
0.643251028, 0.018262707, 0.902121539, 0.808410498, NA, 0.882285296,
0.077569421, 0.249613609, NA, 0.088451014, 0.645309773, 0.626916187,
0.86690534, NA, 0.683221579, 0.486488748, 0.665854791, 0.611034332,
NA, 0.815885005, 0.394817612, 0.403002388, 0.794250738), B = c(0.12246503,
0.557522799, 0.770904525, 0.38075281, 0.290028597, 0.878405308,
0.131643544, 0.211220859, NA, 0.301443669, NA, 0.372821865, 0.076497291,
0.227241864, 0.000343115, 0.65148309, NA, 0.093812247, 0.596673645,
0.018869604, NA, 0.860947689, 0.814111848, 0.17254104, NA, 0.083834541,
0.250760686, 0.705142265, NA), C = c(0.137902549, 0.622560763,
NA, 0.698236443, 0.525661861, 0.645004844, NA, 0.457302727, NA,
0.575188395, 0.343824191, 0.593742731, NA, NA, 0.329546051, 0.267490643,
0.116601451, 0.152577881, NA, NA, 0.472456955, 0.097916066, 0.026085774,
0.212386401, 0.357298987, NA, 0.419539549, 0.768961818, 0.967405211
)), .Names = c("Items", "A", "B", "C"), class = "data.frame", row.names = c(NA, -29L))
#split it up as you suggest
d <- strsplit(d,",")
d <- lapply(d, gsub, pattern=" ", replacement="") #Get rid of trailing spaces
#Convert key to a long data.frame with no NAs
library(reshape2)
key <- melt(key)
names(key)[2] <- "letter" #You might have better name for this
key <- key[complete.cases(key),]
#Extract subsets for each row of data
lapply(d, function(x)key[key$Items %in% x,])