If one end of the stick crosses the event horizon while the other is held by an observer who remains outside the horizon, the stick must break apart. I would say it's easiest to understand this conceptually if you think in terms of a Kruskal-Szekeres diagram for a non-rotating black hole, which has the advantage that light rays are always represented as diagonals 45 degrees from the vertical (unlike in Schwarzschild coordinates, where the coordinate speed of a light ray is not constant), and timelike worldlines always have a slope that's closer to vertical than 45 degrees, so the light cone structure of the spacetime works the same as in Minkowski diagrams from SR (if you're not too familiar with light cones in Minkowski diagrams, see this page). In this coordinate system, the event horizon is actually expanding outward at the speed of light, making it obvious why something that crosses it can never cross back out--it would have to move faster than light! Meanwhile, an observer at a fixed Schwarzschild radius, like the one hovering just above the event horizon, will have a worldline that's a hyperbola bounded from above by the black hole event horizon (it's also bounded from below by a white hole event horizon, but this is just because Kruskal-Szekeres coordinates are defined on the spacetime of an idealized eternal black hole, the white hole event horizon wouldn't be present for a realistic black hole that formed from collapsing matter). This page has a Kruskal-Szekers diagram showing one such hyperbola for an observer hovering at radius r=2.75M in Schwarzschild coordinates, as well as the worldline of an object that falls through the horizon, with light cones drawn at various points along the falling worldline:

There is a useful similarity between this and the Minkowski diagram in SR for a family of accelerating observers, called "Rindler observers" because they have a fixed position in a non-inertial coordinate system known as Rindler coordinates:

The Rindler observers are accelerating in such a way that the distance between them in the instantaneous comoving inertial rest frame of any one of them, at any point on their worldline, is a constant (this type of acceleration is known as Born rigid motion). Since their worldlines are hyperbolas that are bound from above by a worldline which is moving at the speed of light (the dotted line), which can be seen as one side of the future light cone of the point on the diagram where the two dotted lines cross, then since the Rindler observers never enter this future light cone, they can never see light from any event inside it. Thus, the dotted line is a type of horizon for them as long as they continue on the same accelerating path, known as the "Rindler horizon"--see the more detailed discussion on this page.
In your original question, as long as you're dealing with a very large black hole where the tidal forces at the horizon are small, and as long as the stick is fairly short so the region of spacetime where the experiment takes place is very small compared to the Schwarzschild radius, then spacetime will be fairly close to being flat inside that region. Thus, what's seen by the observer hovering at a constant Schwarzschild radius who drops one end of a stick past the event horizon will be similar to what's seen by a Rindler observer in flat spacetime who drops one end of a stick past the Rindler horizon. If the Rindler observer lets one end pass the horizon, but then grabs the other end and exerts enough force on it that it continues to move along with them on the accelerating path, then it's obvious the stick must just split in two, since the end being held by the Rindler observer will never cross the horizon, while the end on the other side of the horizon can't escape back out of it (since that would require it to move faster than light) and can't even maintain a constant distance from it (since that would require it to move at exactly the speed of light).