The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$
where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ were a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.
However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.
The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).
Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.
The big mystery is: why should nature prefer one direction over another? And the answer is still unknown.
From wikipedia: The experiment's purpose was to establish whether or not
conservation of parity (P-conservation), which was previously
established in the electromagnetic and strong interactions, also
applied to weak interactions. If P-conservation were true, a mirrored
version of the world (where left is right and right is left) would
behave as the mirror image of the current world. If P-conservation
were violated, then it would be possible to distinguish between a
mirrored variation of the world and the mirror image of the current
world.
The experiment established that conservation of parity was violated
(P-violation) by the weak interaction. This result was not expected by
the physics community, which had regarded parity as a conserved
quantity.
As opposed to the symmetry expected, in fact 60 % of the gamma rays were emitted in one direction and 40 % in the other. In beta decay, if parity conservation was observed, the electrons should have no particular direction of decay, relative to the nuclear spin of the cobalt atoms. In fact they were (mostly) emitted in a direction opposite that of the nuclear spin.
Conservation of parity
Have a read of the full wiki article with links to concepts behind it, experimental details and the ramifications of her results on the standard model, including time reversal.
BTW, thanks for asking this question, I needed to study it myself this week!
Best Answer
Here's a level scheme for the cobalt-60 decay, from the evaluated nuclear structure data file (ENSDF), with the 99.9% pathway highlighted:
In the beta decay, the spin-parity of the nucleus goes from $5^+\to 4^+$. In the limit where the parity violation is small, parity near-conservation requires the final state to have orbital angular momentum $L=\text{even}$, with $L=0$ strongly preferred. The missing unit of angular momentum must be carried away by the electron and neutrino spins. This is known as a Gamow-Teller transition.
However, in the lepton sector the parity violation is not small. Parity violation is maximal for the antineutrino, whose "north pole" always points in the same direction as its momentum. The electron's "south pole" always points forward in the ultrarelativistic limit, but the electron's polarization is reduced when the electron is at lower energy. To carry away the missing unit of spin, the electron and neutrino "north poles" must both point parallel to the cobalt "north pole." In the cobalt decay path to the nickel excited state, only about $300\,\mathrm{keV}\approx \frac{m_e c^2}{2}$ is available to be shared between the electron and neutrino, so the electron polarization is strong but not overwhelmingly strong. But if you pretend that both the electron and the neutrino are completely polarized, then to carry away the missing unit of angular momentum, their momenta are determined by their spins, with the neutrino traveling along the cobalt's "north axis" and the electron along its "south axis."
Your question is about the electromagnetic transitions in the nickel. Those transitions, with spin-parity $4^+\to 2^+$, are "electric quadrupole (E2)" transitions, where the photon carries away two units of angular momentum but the total parity is unchanged. In that case, the orbital wavefunction is the $L=2$ wavefunction, and the photon angular distribution will go like
$$ Y^\ell_m(\theta,\phi) = Y^2_2(\theta,\phi) = \sqrt{\frac{\text{constant}}{2\pi}} \cdot e^{2i\phi} \sin^2\theta $$
This distribution is strongly peaked in the direction of the cobalt's "equator" at $\theta=0$. That's not a parity-violating distribution, because the "equator" points in the same direction under a parity transformation, while electron preference for the "south pole" would be reversed under a parity transformation.