I suspect that what has confused you is the difference between remaining a fixed distance from the black hole and falling freely into it. Let me attempt an analogy to illustrate what I mean.
Suppose you are carrying a large and heavy backpack. You can feel the gravitational force of the backpack weighing you down. However this only happens because you're staying a fixed distance from the centre of the Earth i.e. you're standing stationary on the Earth's surface. If you and the backpack were to leap from a cliff then (ignoring air resistance) you would feel no gravity as you plummeted downwards and the backpack wouldn't weigh anything.
If we now switch our attention to the black hole, if you attempt to stay a fixed distance from the black hole (presumably by firing the rocket motors on your spaceship) you'd feel the weight of the backpack, and the weight would get bigger and bigger as you approach the event horizon. In fact the weight is given by:
$$ F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$
where $m$ is the mass of the backpack, $M$ is the mass of the black hole, $r_s$ is the event horizn radius and $r$ is your distance from the centre of the black hole. As you approach the event horizon, i.e. as $r \rightarrow r_s$, equation (1) tells us that the force goes to infinity. That's why once you reach the event horizon it is impossible to resist falling inwards.
But you only feel this force because you're trying to resist the gravity of the black hole. If you just fling yourself off your spaceship towards the black hole then you will feel no weight at all. You would fall through the event horizon without noticing anything special. In fact you would see an apparent event horizon retreating before you and you would never actually cross anything that looks like a horizon to you.
But there is another phenomenon that can cause you problems, and this is related to the phenomenon of spaghettification that you mention. At any moment some parts of you will be nearer the centre of the black hole than others. For example if you're falling feet first your feet will be nearer the centre than your head. That means your feet will be accelerating slightly faster than your head, and the end result is that you get slightly stretched. This is called a tidal force, and it happens with all sources of gravity, not just black holes. Even on the Earth the gravitational force on your feet it slightly higher than on your head, though the difference is so small that you'd never notice it.
The thing about a black hole is that because its gravity is so strong the tidal forces can get very strong indeed. In fact they can get so strog that they'd pull you out into a long thin strip like a piece of spaghetti - hence the term spaghettification.
But the tidal forces only become infinite right at the centre of the black hole. They are not infinite at the event horizon, and in fact for large enough black holes the tidal forces at the event horizon can be negligably small. The equation for the variation of gravitational acceleration with distance is:
$$ \frac{\Delta a}{\Delta r} = \frac{c^6}{(2GM)^2} \tag{2} $$
If we take a black hole with the mass of the Sun and use equation (2) to calculate the tidal force we get $\Delta a/\Delta r \approx 10^{9}g$/m. So if you're two metres tall the difference between the acceleration of your head and feet would be $2 \times 10^9g$, where $g$ is the gravitational acceleration at the Earth's surface. This would spaghettify you very effectively. However at the event horizon of a supermassive black hole with the mass of a million Suns the difference between your head and feet would be only 0.001$g$ and you'd struggle to feel it.
I'll try to answer this question based on my intuition about black hole interiors, such as it is, and not any real calculation. A real calculation would be extremely difficult because of the lack of exact solutions describing merging black holes.
Assuming realistic trajectories without enormous acceleration, they probably can't meet. The reason is that your maximum proper time to live once you cross the horizon is comparable to the light crossing time of the black hole (~10 µs per solar mass) while the merging time is probably much larger, depending on how you define it. The two holes that were said to merge in 1/5 of a second had a maximum time to live past the horizon of roughly one millisecond. I don't think you can realistically get from the "leg" to the "hip" region of the interior in that time.
If you treat the star-crossed lovers as classical point particles and allow unlimited acceleration, then I think they can meet, because it's possible to stay arbitrarily close to the horizon after crossing it for an arbitrarily long coordinate (not proper) time. There may be pairs of entry points for which meeting up isn't possible (for example, I think that worldlines that cross a Schwarzschild horizon at precisely antipodal points technically can't meet). But however you want to define "before" and "after" the merger, there should be points that fit your definition for which it is possible.
In the above I assumed that real-world black holes have an unavoidable spacelike singularity like Schwarzschild black holes. Kerr-Newman black holes have interior regions where you can hang out for an arbitrarily long time, and you can definitely meet up in one of those regions. But those regions are not expected to exist in anything but idealized exact vacuum solutions.
Best Answer
First, remember that the notion of an "event horizon" is global statement about the causal structure of spacetime, rather than some local statement. As such talking about the movement of such an object is fraught with difficulty, and any statement about such movement is bound to depend on exactly how one chooses to "slice" spacetime into surfaces of equal "time".
By definition, the event horizon marks the boundary of events in spacetime that can appear in the causal past (past lightcone) of any outside observer. This means that if on any time slice a point lies outside the event horizon, all of the causal past of that point must lie outside of the event horizon.
Now for your intuition. You seem to be led astray by applying Minkowski space intuition to the highly dynamic and strongly curved spacetime around a black hole (merger).
Although I do not necessarily like this language, the simplest way to think about the region of space just "behind" the motion of a black hole is as if this piece of space is being "dragged" along with the black hole. Imagine a ray of light just behind the black hole on a trajectory where it just barely escapes the event horizon. In "reasonable" global coordinates it may appear on a next time slice that this light ray has moved along with the black hole, and now exists at spacial coordinates previously inside the event horizon.