Complexity of recursive function using Master theorem

Assume that the runtime $n$ is measured in the number of elements between i and j. In your method there are two recursive calls

    ...
    int mid = (i + j) / 2;
    if (subsumFun(A, S, i, mid))
        return true;
    ...

and

    ...
    return subsumFun(A, S - s1, mid, j);
}

which both halve the search space, so we have $a = 2$ recursive calls on a problem size of $\frac{n}{b} = \frac{n}{2}$. The work done in

    ...
    if (i >= j)
        return S == 0;
    if (i + 1 == j)
        return A[i] == S || S == 0;
    ...

can assumed to be constant, and the work done in

    ...
    int s1 = 0;
    for (int k = i; k < mid; k++) {
        s1 += A[k];
    }
    ...

is proportional to $\frac{n}{2}$. So the work outside of the recursive calls can be written as $$f(n) = c_1 \frac{n}{2} + c_2$$ for positive constants $c_1, c_2$.

The total runtime of your method can be expressed as the recurrence relation $T$ with

$$T(n) = aT(\frac{n}{b}) + f(n) = 2T(\frac{n}{2}) + c_1 \frac{n}{2} + c_2$$

therefore the Master theorem can be applied:

We have $f \in \Theta(n^{\log_2{2}}) = \Theta(n)$, so using the Master theorem we get $$T \in \Theta(n^{\log_2{2}} \cdot \log{n}) = \Theta(n \cdot \log{n}).$$