Python 3, 89 83 bytes

lambda a,b,c,d:g(*a)+g(*b)+c[0]+"-"+d[2:]
g=lambda c,*s:c in"aeiou"and c or c+g(*s)

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Python 3, 83 81 bytes

lambda a,b,c,d:g(*a)+g(*b)+c[0]+"-"+d[2:]
g=lambda c,*s:c[:c in"aeiou"]or c+g(*s)

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-2 bytes thanks to @JonathanAllan!

Explanation:

g is a recursive function that takes in a splatted string and return the string, truncated after the first vowel.

• lambda c,*s gather the splatted string into the first character c and the rest s.
• c[:c in"aeiou"] or c+g(*s) evaluates to:
  • if c is a vowel, then c[:c in"aeiou"] evaluates to c[:1] which is c. Since c is Truthy, it is returned.
  • if c is not a vowel, then c[:c in"aeiou"] evaluates to c[:0] aka the empty string. The empty string is Falsy, so c+g(*s) (the recursive case) is returned.

lambda a,b,c,d:g(a)+g(b)+c[0]+"-"+d[2:]
g=lambda s:s[:[x in"aeiou"for x in s].index(1)+1]

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Python 3, 89 bytes

lambda a,b,c,d:g(a)+g(b)+c[0]+"-"+d[2:]
g=lambda s:s[:[x in"aeiou"for x in s].index(1)+1]

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Python 3, 89 83 bytes

lambda a,b,c,d:g(*a)+g(*b)+c[0]+"-"+d[2:]
g=lambda c,*s:c in"aeiou"and c or c+g(*s)

Try it online!