Falling into a Schwarzschild black hole in a rocket. Is my description correct? Let's assume the classical Schwarzschild solution with no quantum effects and evaporation.
So, we crossed the event horizon in a rocket.
We find ourselves in a cylindrical space, with the nose of our rocket pointing along the cylinder's axis, that is, along the coordinate $t$. The radius of the cylinder is gradually reducing.
Ahead of us, we see all the matter that fell into the black hole before us, and behind us, we see all the matter that will ever fall after us. But all this matter crossed the horizon at the same time as we did.
But there is no singularity in our space, as we move in space parallel to the singularity, which is separated from us not by space but by time. It is a line, parallel to the cylinder's axis.
Ahead of us, maybe some millions or billions of light-years ahead, there is a star glowing. Since we are so far from it, hardly we can reach it in time before hitting the singularity (is this true?).
People say, there should be some tidal force along the $t$ axis, but I am unsure. Should we experience any destructive forces along $t$? Possibly, yes, but not instantly, although I do not see the source of such force in our space, except the star, which is too far to affect us.
Please, tell me, if this picture correct or not.
P.S. This picture is based on the recent discussion with the user @safesphere, which for some reason has been deleted.
 A: 
We find ourselves in a cylindrical space, with the nose of our rocket pointing along the cylinder's axis, that is, along the coordinate . The radius of the cylinder is gradually reducing.

I would hesitate to call it a cylindrical space because cylinders have a different metric (they are intrinsically flat). But there isn’t really a better word in English, so only the math can really describe the shape. So “cylindrical” is wrong, but the best word I know to convey the gist. Just don’t take that shape too seriously.

Ahead of us, we see all the matter that fell into the black hole before us, and behind us, we see all the matter that will ever fall after us.

We do see all of the previous matter (on our same radial line) at that moment, but we do not see any of the subsequent matter at that moment. We see some of the subsequent matter cross the horizon before we reach the singularity, but there is a lot of matter that crosses the horizon that we can never see cross.

But all this matter crossed the horizon at the same time as we did.

I don’t know of any coordinate chart where this is true. The standard Schwarzschild coordinates don’t cover the horizon, so it isn’t true there. And in charts that do cover the horizon not all crossings happen at the same time.

Ahead of us, maybe some millions or billions of light-years ahead, there is a star glowing. Since we are so far from it, hardly we can reach it in time before hitting the singularity (is this true?).

I assume you are talking about the collapsing star. If so then yes, you can see that from anywhere in the horizon and you will reach the singularity without reaching the star. The distance depends on the size of the black hole.

People say, there should be some tidal force along the  axis, but I am unsure. Should we experience any forces along , destructive to us?

Yes, the spacetime is curved everywhere. You can solve the geodesic deviation equation to show that there are tidal forces everywhere. That said, how destructive they are depends on the size of the black hole. The larger it is the less destructive they are near the horizon, but they become infinite as you get near the singularity.
For a collapsing spacetime the source of the tidal forces is the collapsing star. For the Schwarzschild spacetime there is no source, it is a vacuum spacetime and that is simply one way that a vacuum is allowed to curve.
