The same two pictures also exist in classical mechanics.$^\dagger$ Classical physics can be regarded as a special case of quantum physics in which all observables commute with each other. Just like quantum physics, classical physics can be expressed either in the "Heisenberg picture" or in the "Schrödinger picture", and the two pictures are equivalent: they're just two different ways of thinking about the same thing.
$^\dagger$ As explained in other answers posted here, it's not really a dichotomy, because there are other pictures, too. The point of my answer is that classical physics has the same "pictures" that quantum physics does, not that there are only two pictures.
In classical physics, all observables commute with each other, so we can (and do) always take the state to be an eigenstate of all of the observables.
For this reason, we don't really need to bother distinguishing between the state and the observables in classical physics, but they are logically distinct: the state is what tells us the values of the observables. Observables represent the kinds of things we can measure, and the state tells us what the results of those measurements will be.
(In quantum physics, this distinction is essential, because most observables do not commute with each other, so they cannot all have predictable measurement outcomes. The state tells us what we will get, but only statistically, when we measure observables.)
Two equivalent pictures
To illustrate the two pictures in classical mechanics, consider the classical mechanics of a system of objects interacting with each other as in Newton's model of gravity:
In the Heisenberg picture, the observables are the fact that we can measure the objects' positions at any desired time, and those positions are related to each other by the equations of motion. The state specifies a particular solution of the equations of motion, which endows all of those observables (at each time) with specific values.
In the Schrödinger picture, the observables are the fact that we can measure the object's positions and momenta. The state specifies a particular set of values for the positions and momenta at any given time, and the time-evolution of the state tells us how the positions and momenta evolve in time.
If the distinction seems insignificant, that's because the two pictures are indeed equivalent. Either one by itself accounts for all of the time-dependence. In classical physics, we subconsciously switch back and forth between these two equivalent pictures. After enough experience, we do this subconsciously in quantum physics, too.
The example in the question
Regarding the ladder-operators example that is mentioned in the question: I didn't watch the lecture, but the notation $\alpha(t)$ indicates that the lecturer is working in the Heisenberg picture. The operator $\alpha(t)$ is a time-dependent observable. The Heisenberg equations of motion are probably Maxwell's equations (just guessing because I didn't watch the lecture), but with operator-valued components $\alpha(t)$ of the fields.
Even though the observable is time-dependent (a different operator at different times), the fact that the time-dependence is governed by the Heisenberg equation of motion implies that we can write all of these observables in terms of a common set of operators, the ladder operators. In this context, the ladder operators are not associated with any particular time. They're just operators on the Hilbert space that we can use to express all of the observables $\alpha(t)$.
The classical-physics analogue is that the general time-dependent electromagnetic field that satisfies Maxwell's equations can be written in terms of a fixed set of unspecified coefficients. The time-dependent components of the electromagnetic field are the observables. The state selects a specific solution by specifying the values of the coefficients in that general solution. The ladder operators are analogous to the coefficients, except that we can't completely "specify their values" in quantum physics, because they don't commute with each other.