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How to do I convert date format from "09-12:36:10" into "09-Month-Year 12:36:10", replacing “Month” with the current month name (three-letter abbreviation) and “Year” with the current year (four-digit), using awk?
For example, change "09-12:36:10" to "09-JUL-2018 12:36:10".

simply,

echo 09-12:36:10 | awk -F '-' -v date="$(date +"-%b-%Y " )" ' print $1date$2'

If you are keen to have month in upper case:

echo 09-12:36:10 | awk -F '-' -v date="$(date +"-%b-%Y " )" ' print toupper($1date$2)'

Not sure how robust you are needing this, but here is a start:

#!/usr/bin/awk -f
BEGIN 
 FS = "-"


 print $1 "-MONTH-YEAR", $2

If you have GNU Awk (gawk), you can use its built-in systime and strftime

gawk -F'[-]' 'print $1, strftime("%b %Y", systime()), $2' 

Ex.

$ echo '09-12:36:10' | gawk -F'[-]' 'print $1, strftime("%b %Y", systime()), $2'
09 Jul 2018 12:36:10

If you need the abbreviated month in upper case, try %^b e.g.

$ echo '09-12:36:10' | awk -F'[-]' 'print $1 strftime("-%^b-%Y", systime()), $2'
09-JUL-2018 12:36:10