Which Ninjutsu am I casting?

JavaScript (ES6), 95 bytes

A shorter version suggested by @tsh

Expects [1,2,3] for "TCJ".

(a,b,c)=>a^b&&a^c&&b^c?',Hu,Do,Sui,Hyo,Rai,Ka'.split`,`[c||7-b]+'ton':b?'Bunny':'Fuma Shuriken'

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---

JavaScript (ES6), 99 bytes

Expects "012" for "TCJ".

s=>/(.).?\1/.test([,b,c]=s)?'Bunny':b?'Hu,Do,Sui,Hyo,Rai,Ka'.split`,`[c||5-b]+'ton':'Fuma Shuriken'

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Commented

s =>                         // s = input string
  /(.).?\1/                  // regex to detect a duplicate character
  .test([, b, c] = s)        // apply it to s; at the same time, load the 2nd
                             // and 3rd characters into b and c respectively
  ?                          // if there's a duplicate:
    'Bunny'                  //   return 'Bunny'
  :                          // else:
    b ?                      //   if the 2nd character is defined:
      'Hu,Do,Sui,Hyo,Rai,Ka' //     lookup string of prefixes
      .split`,`              //     split it
      [c || 5 - b]           //     use c if it's defined, or 5 - b otherwise
      + 'ton'                //     append the suffix
    :                        //   else:
      'Fuma Shuriken'        //     return 'Fuma Shuriken'

R, 153 149 144 139 bytes

Edit: -5 bytes thanks to Giuseppe

function(x)`if`(any(table(x)>1),'Bunny',`if`((l=sum(x|1))-1,a[l*3-6+x[l]],'Fuma Shuriken'));a=paste0(scan(,''),'ton')
Ka
Rai
Hyo
Hu
Do
Sui

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Input is 1 for 'Ten', 2 for 'Chi', 3 for 'Jin'.

Ungolfed:

a=paste0(scan(,''),'ton')       # scan(,'') reads the subsequent strings until it finds an empty one,
Ka                              # paste0 joins them each together with 'ton'.
Rai
Hyo
Hu
Do
Sui

function(x){
 if(any(table(x)>1))'Bunny'     # if there are any duplicates: 'Bunny'
 else{                          # otherwise
  l=length(x)                   # define l=length of x (the R golfy way: sum(x|1) because 'length' is 6 letters)
  if(l-1)a[l*3-6+tail(x,1)]     # if it isn't length 1, output the relevant element from the vector 'a' of ninjutsu names 
  else'Fuma Shuriken'           # otherwise (it's length 1): 'Fuma Shuriken'.
 }
}
```

Haskell, 116 109 bytes

saved 7 bytes thanks to @ovs, and obscure things in the stdlib (why is word but not split in Prelude?)

n[_]="Fuma Shuriken" -- matches a list of one element
n(a:b)|a`elem`b="Bunny" -- matches a list starting with a where a is in b `| `(cond)` =` is a guard that checks cond
n[a,b]=words"Ka Rai Hyo Ho Do Sui"!!b++"ton" -- !! is indexing, ++ is appending
n[a,b,c]=n[b+3,c+3] -- don't mistake n[] as indexing, it's calling n with a list.
-- b+3 so it will still catch duplicates where b=c

Highly readable (IMO), and highly not PHP (sorry PHP). This makes excessive use of pattern matching. It takes input as a list of Ten=0, Chi=1, Jin=2. Trailing whitespace and comments (starting with --) are not counted.

ideone it!