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.
It might be easiest to open your idea of coordinates to include lots and lots of different coordinates.
For instance if you took a regular Schwarzschild solution then you can imagine putting almost any kind of coordinate system on it. To make paint a concrete picture lets first draw the solution in a Kruskal-Szekeres coordinate system.
In a Kruskal-Szekeres coordinate system you can draw the $y$ axis as vertical and the $x$ axis as horizontal. And for fixed $b\geq 0$ there is are hyperbola like $y=+\sqrt{x^2+b^2}$ that correspond to surfaces of constant areal coordinate inside the black hole event horizon (it is the event horizon when $b=0$). And there is one value of $b$ that corresponds to the singularity.
Similarly, for fixed $b\geq 0$ there is are hyperbola like $y=-\sqrt{x^2-b^2}$ that correspond to surfaces of constant areal coordinate inside the white hole event horizon (it is the event horizon when $b=0$). And there is one value of $b$ that corresponds to the singularity.
And for fixed $b\geq 0$ there is are hyperbola like $x=+\sqrt{y^2+b^2}$ that correspond to surfaces of constant areal coordinate outside the black hole event horizon (it is the event horizon when $b=0$). And there is no value of $b$ that corresponds to the singularity (assuming the mass of the black hole is positive) because you are outside.
Similarly, for fixed $b\geq 0$ there is are hyperbola like $x=-\sqrt{y^2+b^2}$ that correspond to surfaces of constant areal coordinate outside the black hole event horizon in another universe (it is the event horizon when $b=0$). And there is no value of $b$ that corresponds to the singularity (assuming the mass of the black hole is positive) because you are outside.
That's how the the hyperbolas in the Kruskal-Szekeres coordinates relate to the areal coordinate. And the areal coordinate is the one people sometimes call radial, but only in the outside part far from the horizon does it start to become close to a radial distance. Really it's a surface such that if you keep it fixed at $r$ and keep Schwarzschild time fixed at $t$ you get a surface of area $4\pi r^2.$ So it is an area-al (areal) coordinate, not a radius-al (radial) coordinate.
Next we look at Schwarzschild time $t$. For a fixed $a\in[-1,+1]$ then outside there are curves like $y=ax$ that correspond to curves of constant Schwarzschild time with $a=0$ corresponding to $t=0$ and $y=0.$ And positive $a$ corresponding to positive Schwarzschild time, and negative $a$ corresponding to negative Schwarzschild time.
As you go from $a=0$ to $a=+1$ you go from $t=0$ to $t=+\infty$ and through every positive $t$ in between. And $a=1$ is $t=+\infty$ where $t$ is Schwarzschild time and $y=1x$ is part of the event horizon.
Similarly, as you go from $a=0$ to $a=-1$ you go from $t=0$ to $t=-\infty$ and through every negative $t$ in between. And $a=-1$ is $t=-\infty$ where $t$ is Schwarzschild time and $y=-1x$ is the other part of the event horizon.
Note that all of those lines only cover the outsides of the event horizon ($a\in[-1,+1]$)./To cover the inside you can again draw a series of lines but in this region you still pick an $a\in[-1,+1]$ but the line is the line $x=ay$ and again $a=+1$ is $t=+\infty$, $a=0$ is $t=0$ and $a=-1$ is $t=-\infty$ but inside the event horizon, the Schwarzschild time $t$ is not a timelike coordinate. In particular, it is increasing $y$ that is future directed. So for any point in the event horizon of the balck hole there is a future pointing null ray with increasing $y$ and decreasing $x$ (ingoing future pointing) and another one with increasing $y$ and increasing $x$ (outgoing future pointing). The former has $t$ getting smaller, the latter has $t$ getting larger. Both have the areal coordinate getting smaller.
You can see this because null rays go at 45 degrees in a Kruskal-Szekeres coordinate system. And so the left-and-up going one passes through lines like $x=ay$ with different (and decreasing) $a$ that's the ingoing null ray. And the right-and-up going one passes through lines like $x=ay$ with different (and increasing) $a$ that's the outgoing null ray.
So far I haven't breathed a word about apparent horizons. I'm just drawing a picture of the Schwarzschild solution in the Kruskal-Szekeres coordinate system and making sure you see where the curves of constant Schwarzschild time $t$ are (they are the lines through the origin). And where the curves of constant areal coordinate are (they are hyperbolas like $y=+\sqrt{x^2+b^2}$ or $x=+\sqrt{y^2+b}.$
And we briefly identified different curves as future/past pointing and as ingoing/outgoing.
increasing $y$ and decreasing $x.$ Don't gloss over a single bit of this. Nothing is deep. Nothing is complicated. But getting it wrong means we'd just talk past each other, so it is essential that you learn it so that we can communicate.
So now we can be inside or outside and use the Kruskal-Szekeres coordinates or the Schwarzschild time $t$ areal coordinate $r$ and the angles $\theta$ and $\phi.$ But we can make lots of other coordinates systems besides those two. It doesn't change which curves are ingoing, outgoing, future pointing, or past pointing. All of that is given by the Kruskal-Szekeres coordinates becasue it was designed to respect causality.
Any event located inside the black hole event horizon has ingoing and outgoing future pointing null directions (going left-up in Kruskal-Szekeres and going right-up in Kruskal-Szekeres). Inside the event horizon you can make alternative coordinates where one of these null rays can have increasing coordinate value for a bit, even though eventually it comes back.
But those range of coordinates never cover the outside. For instance if you used Schwarzschild time and areal coordinate then the areal coordinate never gets larger than the Schwarzschild radius and the Schwarzschild time never reaches the event horizon either, no matter what finite value it reaches in either direction. So it's a fine coordinate chart for the inside. But do keep in mind that inside the Schwarzschild time is a spacelike coordinate, and the areal coordinate is a timelike coordinate.
But it's just one coordinate system. And there are lots. And you can pick ones where the null rays gets a larger value for a while and then reaches a maximum. Lots of choices. There are even choices of coordinates where there are no apparent horizons.
So you might be used to Schwarzschild coordinates where the ingoing null ray has Schwarzschild time strictly decrease and the outgoing null ray has Schwarzschild time strictly increase and both have areal coordinate strictly decrease (because they are future pointing rays). But in other coordinates the coordinates of those same rays could increase for a while then reach a maximum, then decrease.
Why is it useful? Since it is local (a null ray can get a larger coordinate, a maximum coordinate, then a smaller one) it's easier to find. And if you know it is inside an event horizon, then now you know that somewhere outside that surface is an event horizon and those were hard to find.
So you might want to find an event horizon and it's hard to find. So you try some coordinates and find an apparent horizon. Now you know there is an event horizon and you narrowed down where it is.
That's nice if for instance you are counting on cosmic censorship to hide a singularity and you are doing numerical relativity and want to avoid a singularity, because the outside of the event horizon (what you want to solve for) is outside the apparent horizon, so you know now you've got more than enough to have what you want.
Sure it's a bit of overkill to solve for everything outside the apparent horizon, but if you do that, then you know you got everything you needed.
Best Answer
That's a subtle question because it depends on what you mean by energy.
Suppose we take some object a long way from the black hole and let it fall. As it falls it's speed increases (obviously) and therefore it's kinetic energy increases. However if you are sat outside the black hole watching the object you'll see something strange as the object nears the event horizon. You'll see the object start to slow as it approaches the horizon and indeed at the horizon it will come to a complete stop. This is the source of the numerous questions asking whether anything can really fall into a black hole e.g. this and this. Incidentally while searching for such questions I found the question A Conflict with Black Holes that is relevant to your question.
So from our perspective outside the black hole the kinetic energy of the body rises to a maximum then falls again as it nears the horizon.
However there is more to this. Suppose you jump on a powerful rocket and hover at some distance $d$ from the event horizon, then you measure the velocity at which the object passes you. As you approach the horizon, i.e. as $d \rightarrow 0$, the velocity the object passes you will tend to $c$ and therefore it's energy will tend to infinity. However you cannot hover at the event horizon (that would take an infinitely powerful rocket) so no observer will ever see the velocity become $c$ and the kinetic energy become infinite.
The other option you might take is to jump into the black hole alongside the object so you are falling with it. This is actually the least interesting option (until you hit the singularity and the associated spectacularly messy death of course) because you will simply see the object float, weightless, beside you and you won't measure any change in it at all.