What is the time complexity of binary sum (recursion)
First of all, calculating the sum can't be better than $O(n)$ in the general case because you have to inspect all elements in order to determine the sum.

The function also does not halve the problem size every step: instead of choosing one subarray, as in binary search, we sum both subarrays. This does not save work at all.

We can determine the running time using the recurrence relation:

$$T(0) = T(1) = 1$$
$$T(n) = 2 \times T(n/2) + 1$$

Setting $k = log_2(n)$ we can write this this as:

$$T'(0) = 1$$
$$T'(k) = 2 \times T'(k-1) + 1$$

Now we can use:

$$T'(n) = \sum_{i = 1}^{k} 2^i \times 1$$
$$= 2^{k + 1} - 1 = 2 \times 2^{k} - 1$$
$$= 2 \times n - 1$$