Note: this gives a plausible explanation, not a whole solution.

Hi, the BSD-quotient is $L(E_n,1)/\Omega_n = Sha\cdot\prod_p c_p(E_n)/|T|^2$. I think your question is mostly equivalent to asking why this quotient is an integer when twisting by $n$? The torsion is of size 4 (maybe there is an exceptional twist where it is larger, you did not specify). I think because of the full 2-torsion, the Tamagawa number should be 4 at any prime dividing $n$, except those dividing the original conductor. So if $n$ has at least two prime factors coprime to $N$ it seems you are done. When twisting by a prime or some $n$ with nontrivial gcd with $N$, then you need to take a more careful analysis, but the original curve itself should already have nontrivial Tamagawa numbers at some prime if indeed it has full 2-torsion. In short, the product of Tamagawa numbers outweighs the torsion in the denominator.

The definition of "0" here is just the cusp 0 on the boundary of the complex upper half plane. One is essentially integrating the modular form as $\int_0^{i\infty} f(z) dz$, to get the modular parametrization. But there are the constants like $2\pi$ and the Manin constant to worry about.