The detection of 2 light pulses by Herman occur at the same time, and at the same place. Let's call it:
$$ E_3 = (0, 0)_{\rm Herm} $$
This is one event. An event is a single point in spacetime.
I can transform it into any frame (a primed frame, $S'$), and it will be:
$$ E_3' = (t', x')_{S'} $$
That is a single event, period. It must be simultaneous and co-located. It is by definition.
If we go to Al's frame, it can be:
$$ E_3 = (0, 0)_{\rm Al} $$
where, as with Herman, I used it to define the origin of the coordinates.
For there to be a discrepancy with simultaneity, we need a spatial separation of two distinct events.
That would fall on the emission of the light. In Hermann's frame, those events occur at:
$$ E_1 = (-L/2, -L/2)_{\rm Herm} $$
$$ E_2 = (-L/2, +L/2)_{\rm Herm} $$
where $L$ is the length of car, and $c=1$. We see that $t_1=t_2= -L/2$; the emission is simultaneous.
They cannot be simultaneous in Al's frame. A Lorentz Transformation by $-v$ shows:
$$ E_1 = \big(\gamma(-L/2-(-vL/2)), \gamma(-L/2-v(-L/2)\big)_{\rm Al}$$
$$ E_1 = \big(-\frac 1 2 \gamma L(1+v), -\frac 1 2 \gamma L(1+v)\big)_{\rm Al} $$
$$ E_2 = \big(\gamma(-L/2-(+vL/2)), \gamma(+L/2-v(-L/2)\big)_{\rm Al}$$
$$ E_2 = \big(-\frac 1 2 \gamma L(1-v), \frac 1 2 \gamma L(1-v)\big)_{\rm Al} $$
So we see that for Albert, $E_1$ occurs before $E_2$, that is, the trailing photon is emitted 1st and has to catch up with Hermann, and the leading photon is emitted second, with Hermann running into it. Hence the $\gamma(1\pm v)$ difference.