You can find the first statement, for instance, in van der Waerden, Modern Algebra I, Section 61. In particular, under the inclusion of the Galois group of the reduction in the Galois group of the original polynomial, the cycle structure matches up.

For the statement when you drop the separability assertion, the answer is no. Here is an example in which the cycle structure of the two groups are incompatible. While this does not completely settles your second question, it might convince you that the answer is "no".

The polynomial f=x4+x2+9 is irreducible over $\mathbb{Q}$ and has discriminant equal to $420^2$. Therefore the Galois group of $f$ is contained in the alternating group of order 4 and contains no transposition. The reduction of the polynomial modulo 3 has Galois group that is a transposition. Even though the Galois group of $f$ contains elements of order two, the cycle structure do not match up.

You can find the first statement, for instance, in van der Waerden, Modern Algebra I, Section 61. In particular, under the inclusion of the Galois group of the reduction in the Galois group of the original polynomial, the cycle structure matches up.

For the statement when you drop the separability assertion, the answer is no. Here is an example in which the cycle structure of the two groups are incompatible. While this does not completely settle your second question, it might convince you that the answer is "no".

The polynomial $f=x^4+x^2+9$ is irreducible over $\mathbb{Q}$ and has discriminant equal to $420^2$. Therefore the Galois group of $f$ is contained in the alternating group of order 4 and contains no transposition. The reduction of the polynomial modulo 3 is $x^2(x^2+1)$ and therefore has Galois group that is a transposition. Even though the Galois group of $f$ contains elements of order two, the cycle structures do not match up.