Well, the bottom line is that it is equivalent.
The intermediate mod's do not modify the value of the expression; this is because all the arithmetic done by the expression (except for the calculation of the exponent (a + u * x)) is done modulo N; it is always safe to do intermediate reductions.
And, in this case, it's actually necessary; g^x is quite a huge number, and you really have to compute it modulo N, the actual value is just too big.
As for why B+ and N- instead of B-, well, it is likely that the modular exponentiation primitive that the programmer used wasn't able to handle negative bases. By reducing the value of k * g^x modulo N (which is safe, as above), that value is always less than N; and so N - (k * g^x % N) is always positive, and so is B + (N - (k * g^x % N)). Also note that this modification is also safe, because for all $A$, $B$, $N$, we have:
$A - B = A + N - B = A + (N - B) \mod N$
I would note that there are some further reductions that could have been done:
The exponent could have been computed modulo (N-1), as in (a + u*x) % (N-1). This is a safe reduction because N is prime (and Fermat's Little Theorem); this would speed up the second modular exponentiation operation by giving it a smaller exponent. Note that the modular exponentiation primitive cannot do it (it couldn't assume N is prime).
When computing (B + (N - ((k * (g^x % N)) % N)), they could have done a final % N to it; this might speed up the second modular exponentiation operation (by giving it a smaller base). However, unlike the above operation, the code that computes the modular exponentiation might do that already (because that is always safe, whether or not N is prime).
