How to extract patterns from a file and fill an bash array with them?

fill an array with all the ocurrences of the pattern

First convert your file to have meaningful delimiter, ex. null byte, with ex. GNU sed with -z switch:

sed -z 's/"\([^"]*\)"[^"]*/\1\00/g'

I've added the [^"]* on the end, so that characters not between " are removed.

After it it becomes more trivial to parse it.

You can get the first element with:

head -z -n1

Or sort and count the occurrences:

sort -z | uniq -z -c

Or load to an array with bash's maparray:

maparray -d '' -t arr < <(<input sed -z 's/"\([^"]*\)"[^"]*/\1\00/'g))

Alternatively you can use ex. $'\01' as the separator, as long as it's unique, it becomes simple to parse such data in bash.

Handling such streams is a bit hard in bash. You can't set variable value in shell with embedded null byte. Also expect sometimes warnings on command substitutions. Usually when handling data with arbitrary bytes, I convert it with xxd -p to plain ascii and back with xxd -r -p. With that, it becomes easier.

The following script:

cat <<'EOF' >input
"My name
is XXX"
"My name is YYY"
"Today
is
the "
EOF

sed -z 's/"\([^"]*\)"[^"]*/\1\x00/g' input > input_parsed

echo "##First element is:"
printf '"'
<input_parsed head -z -n1 
printf '"\n'

echo "##Elemets count are:"
<input_parsed sort -z | uniq -z -c

echo
echo "##The array is:"
mapfile -d '' -t arr <input_parsed
declare -p arr

will output (the formatting is a bit off, because of the non-newline delimetered output from uniq):

##First element is:
"My name
is XXX"
##Elemets count are:
      1 My name
is XXX      1 My name is YYY      1 Today
is
the 
##The array is:
declare -a arr=([0]=$'My name\nis XXX' [1]="My name is YYY" [2]=$'Today\nis\nthe ')

Tested on repl.it.

This may be what you're looking for, depending on the answers to the questions I posted in a comment:

$ readarray -d '' -t arr < <(grep -zo '"[^"]*"' file)

$ printf '%s\n' "${arr[0]}"
"My name
is XXX"

$ declare -p arr
declare -a arr=([0]=$'"My name \nis XXX"' [1]="\"My name is YYY\"" [2]=$'"Today\nis\nthe "')

It uses GNU grep for -z.

Sed can extract your desired pattern with or without newlines. But if you want to store the multiple results into a bash array, it may be easier to make use of bash regex.
Then please try the following:

lines=$(< "file")                   # slurp all lines
re='"[^"]+"'                        # regex to match substring between double quotes
while [[ $lines =~ ($re)(.*) ]]; do
    array+=("${BASH_REMATCH[1]}")   # push the matched pattern to the array
    lines=${BASH_REMATCH[2]}        # update $lines with the remaining part
done

# report the result
for (( i=0; i<${#array[@]}; i++ )); do
    echo "$i: ${array[$i]}"
done

Output:

0: "My name
is XXX"
1: "My name is YYY"
2: "Today
is
the "