JavaScript (ES6), 108 107 bytes

let f =

a=>[4,...a].map(d=>r[p+=[-99,-98,-97,-1,1,97,98,99][d]]=String.fromCharCode(i++),r=[],i=65,p=i*i)&&r.join``

console.log(f([4,7,1,6,6,6,0,3,1]));

Octave, 145 138 131 123 105 103 90 87 85 bytes

@(a){[~,k]=unique(cumsum([1 fix((z=a+(a>3))/3);1 mod(z,3)]'-1),'rows');[k'+64 '']}{2}

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Thanks to Suever 2 bytes saved!

Previous answer 103 bytes:

@(a)[nonzeros(accumarray(cumsum([1 fix((z=a+(a>3))/3);1 mod(z,3)]'-1)+30,65:nnz(++a)+65,[],@max)')' '']

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First try 145 byts

@(a){[x y]=find(~impad(1,1,1));c=cumsum([0 0;([y x]-2)(++a,:)]);c=c-min(c)+1;n=nnz(a);[nonzeros(sparse(c(:,1),c(:,2),65:n+65,'unique')')' '']}{5}

Some Explanations

@(a){
    [x y]=find([1 0 1]|[1;0;1]);                            %generate 2d coordinates corresponding to 1d input indices
    XY = [y x]-2;
    c=cumsum([0 0;XY(++a,:)]);                              %cumulative sum of coordinates to find position of characters
    c=c-min(c)+1;n=nnz(a);
    [nonzeros(sparse(c(:,1),c(:,2),65:n+65,'unique')')' ''] %using sparse matrix to place characters at specified positions
    }{5}

MATL, 64 58 57 50 46 40 37 bytes

nQ:64+cPFF3Y62#fwh2-GQY)&vYsPHH$6#Xu)

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Explanation

        % Implicitly grab the input
nQ      % Determine the number of inputs and add 1 (N)
:       % Create an array from [1...N]
64+     % Add 64 (converts this to ASCII codes [65....65+N])
c       % Convert from ASCII codes to actual characters (['A'...Nth_letter])
P       % Reverse the string
FF      % Push the array [0, 0] to the stack
3Y6     % Push the 2D matrix [1 1 1; 1 0 1; 1 1 1] to the stack
2#f     % Get the rows, columns of the 1's in the previous matrix
wh      % Horizontally concatenate these row/column indices
2-      % Subtract two from each of them so that we get values between -1 and 1
GQY)    % Grab the input, add one (1-based indexing) and index into this row/col matrix
&v      % Vertically concatenate the [0 0] with this new matrix
Ys      % Compute the cumulative sum down the columns which yields 2D coordinates for each
        % successive letter
P       % Flip the result (this is to work around a bug in MATL)
HH$6#Xu % Compute the unique rows (sorted) and find the first occurrence of each unique row
)       % Use this to index into the initial character array. Since we flipped
        % the initial array of letters and the input to the unique function we
        % will get the LAST letter to occur at each point
        % Implicitly display the result

Python 2, 180 178 bytes

def f(d,a=[[""]*26 for _ in[1]*26],x=0,y=0,j=1):
 a[0][0]="A"
 for i in d:y+=(i>4)-(i<3);x+=(`i`in'247')-(`i`in'035');a[y][x]=chr(j+65);j+=1
 return"".join("".join(i)for i in a)

MATLAB, 87 bytes

function a=f(s);i='%&''?AYZ['-64;[~,a]=unique([0 cumsum(i(s+1))],'last');a=char(a'+64);

PHP, 125 bytes

function g($d){$r[0]=chr(65);foreach($d as$i=>$v)$r[$c+=[-27,-26,-25,-1,1,25,26,27][$v]]=chr($i+66);ksort($r);echo join($r);}

Run it like this:

g([4,7,1,6,6,6,0,3,1]);

If strings are OK as input, then this is 121 bytes and will work on the command line:

$r[0]=chr(65);for(;($n=$argv[1][$i])!=null;)$r[$c+=[-27,-26,-25,-1,1,25,26,27][$n]]=chr($i+++66);ksort($r);echo join($r);