Note: In the question's current form, a sed solution is feasible - this was not the case originally, where the last ;-separated field of the joined lines needed transforming as a whole, which prompted the awk solution below.
Joining lines 7 and 8 as-is, merely by removing the line break between them, can be achieved with this simple sed command:
sed '7 { N; s/\n//; }' file.csv
awk solution:
awk '
BEGIN { FS = OFS = ";" }
NR==7 { r = $0; getline; sub(/^"",,/, ""); $0 = r $0 }
1
' file.csv
Judging by the OP's comments, an additional problem is the presence of CRLF line endings in the input. With GNU Awk or Mawk, adding RS = "\r\n" to the BEGIN block is sufficient to deal with this (or RS = ORS = "\r\n", if the output should have CRLF line endings too), but with BSD Awk, which only supports single-character input record separators, more work is needed.
BEGIN { FS = OFS = ";" } tells Awk to split the input lines into fields by ; and to also use ; on output (when rebuilding the line).
Pattern NR==7 matches input line 7, and executes the associated action ({...}) with it.
r = $0; getline stores line 7 ($0 contains the input line at hand) in variable r, then reads the next line (getline), at which point $0 contains line 8.
sub(/^"",,/, "") then removes substring "",, from the start of line 8, leaving just 8-9012.
$0 = r $0 joins line 7 and modified line 8, and by assigning the concatenation back to $0, the string assigned is split into fields by ; anew, and the resulting fields are joined to form the new $0, separated by OFS, the output field separator.
Pattern 1 is a common shorthand that simply prints the (possibly modified) record at hand.
