Efficient Typing on a Game Boy

JavaScript (ES6), 147 bytes

s=>s.replace(/./g,c=>(q=p,p="AHOVBIPWCJQXDKRYELSZFMY.GNU ".indexOf(c),"<<<>>>".substring(3,((p>>2)+10-(q>>2))%7)+["","v","vv","^"][p-q&3]+"."),p=0)

An interesting behaviour of substring is that it exchanges the arguments if the second is less than the first. This means that if I calculate the optimal number of left/right presses as a number between -3 and 3, I can add 3, and take the substring of <<<>>> starting at 3 and I will get the correct number of arrows. Meanwhile the down/up presses are simply handled by looking up an array using a bitwise and of the difference in rows with 3; this way is slightly shorter as there are fewer array elements.

Dyalog APL, 61 bytes

4 7∘{∊'.',⍨⍉↑b⍴¨¨'^v' '<>'⌷¨⍨⊂¨a>b←a⌊⍺-a←⍺|↓2-/0,⍺⊤⍵⍳⍨⎕a,'.'}

assumes ⎕IO←0

⎕a,'.' the alphabet followed by a full stop

⍵⍳⍨ find the argument's chars there as indices 0..26 (' ' and all others will be 27)

⍺⊤ encode in base 7 (note the left arg ⍺ is bound to 4 7), get a 2×n matrix

0, prepend zeros to the left

2-/ differences between adjacent columns

↓ split the matrix into a pair of vectors

a←⍺| take them modulo 4 and 7 respectively, assign to a

b←a⌊⍺-a make b the smaller of a and its modular inverse

'^v' '<>'⌷¨⍨⊂¨a>b choose ^ or v for the first vector and < or > for the second, based on where a differs from b

b⍴¨¨ repeat each of those b times

⍉↑ mix the two vectors into a single matrix and transpose it, get an n×2 matrix

'.',⍨ append .-s on the right

∊ flatten

Ruby, 107 bytes

->s{c=0
s.tr(". ","[\\").bytes{|b|b-=65
print ["","^","^^","v"][c/7-b/7],(d=(c-c=b)%7)>3??>*(7-d):?<*d,?.}}

Ungolfed in test program

f=->s{                                 #Input in s.
  c=0                                  #Set current position of pointer to 0.
  s.tr(". ","[\\").                    #Change . and space to the characters after Z [\
  bytes{|b|                            #For each byte b,
    b-=65                              #subtract 65 so A->0 B->1 etc.
    print ["","^","^^","v"][c/7-b/7],  #Print the necessary string to move vertically.
    (d=(c-c=b)%7)>3?                   #Calculate the horizontal difference c-b (mod 7) and set c to b ready for next byte.
       ?>*(7-d):?<*d,                  #If d>3 print an appropriate number of >, else an appropriate number of <.
    ?.                                 #Print . to finish the processing of this byte.
  }
}

#call like this and print a newline after each testcase
f["FLP.TKC"];puts  
f["MOYLEX"];puts   
f["FEERSUM"];puts  
f["MEGO"];puts     
f["A CAT"];puts    
f["BOB"];puts