The answer should be {1, 1, 1, 1, 0, 0, 0, 0 } ;

num3 = {1, 1, 1, 1, 0, 0, 0};
num4 = {1, 1, 1, 1, 0, 0, 0};
IntegerDigits[FromDigits[num3, 2] + FromDigits[num4, 2], 2]

In your code use Floor instead of Round. And AppendTo puts the first result into Position 1 of your result and so on: use Reverse

num3 = {1, 1, 1, 1, 0, 0, 0};
num4 = {1, 1, 1, 1, 0, 0, 0};
add = {};
acc = 0;
For[i = Length[num3], i > 0, i--, 
 AppendTo[add, Mod[num3[[i]] + num4[[i]] + acc, 2]];
 acc = Floor[(num3[[i]] + num4[[i]] + acc)/2];
 ]
Reverse@AppendTo[add, acc]

By the way: you should avoid AppendTo and use other methods.

And one could do it like this (without explicit For or While ).

Define

add1[{a_, b_}] := Module[{xx},
  arg = Transpose[Reverse /@ {a, b}] /. {1 -> True, 0 -> False};
  e1 = Reverse /@ 
    Transpose[Map[Boole, ({Xor @@ #, And @@ #} & /@ arg), {2}]];
  If[e1[[2, 1]] == 1, ovflg = 1];
  xx = RotateLeft[Join[e1[[2]], {0}]];
  {e1[[1]], Drop[xx, -1]}
  ]

and

add[a_, b_] := Module[{n, xx},
ovflg = 0;
n = Length[a];
If[n != Length[b], Return["error"]];
xx = Nest[add1, {a, b}, n - 1];
xx[[1]]]

Then

a = {1, 0, 1, 1, 1};
b = {0, 1, 1, 0, 0};
add[a, b]
ovflg

or

add[{0, 1, 1, 1}, {0, 1, 0, 1}]
ovflg

Thanks for the suggestion, it works now.

Thank you for this translation! I was stuck at the indexing part and hence, wasn't able to write the next part. I will check this out, and check for ways to do it without For and AppendTo.