Consider the state space representation of the following discrete time system: $$ x_{k+1} = A x_k + B x_k \\ y_{k} = C x_k $$
The system is observable if the row rank of the observability matrix $\mathcal{O}$ is equal to $n$ where $n$ is the number of state variables. The observability matrix is given by
$$ \mathcal{O} = \begin{bmatrix} C \\ CA \\ CA^2 \\ \vdots \\ CA^{n-1} \end{bmatrix} $$
However, if $C$ is of rank $n$, then the system is observable, so the additional terms need not be evaluated. Similarly, if $\begin{bmatrix} C \\ CA \end{bmatrix}$ is observable, then additional terms need not be evaluated. This sort of described how many measurements are required to determine the states uniquely. However, it is known that if $\mathcal{O}$ is not rank $n$, then additional steps will not allow for determining the states uniquely; thus, I believe that is why the observability matrix is defined in this way.
So, I am stuck wondering, does a definition exist for a $K$-step observability? This concept is useful but I am having trouble finding any sources discussing a notion of $K$-step observability or something similar.
This doesn't directly answer your question regarding K-steps observability, but this would be too long for a comment. I would first like to note that observability is a Boolean (either true or false) property of a state space model. There are also other ways of showing this, such as the Hautus test or observability Gramian. This Hautus test doesn't have any connection to the minimal number of samples required in order to be able to reconstruct the full state. For the observability Gramian one could also take a finite summation. This would also be required in order for the Gramian to remain bounded when $A$ has observable unstable modes. It can be noted that such finite time Gramian would have the same rank as the K-steps observability matrix, since that Gramian is equal to $\mathcal{O}_K^\top \mathcal{O}_K$. But the infinite time Gramian can also be found by solving a Lyapunov equation.
The only place I can think of where one might use this K-steps observability would be for a deadbeat observer. Such observers would be very sensitive to noise in the measurements, even if $C$ would have rank $n$, and in practice measurements always have some noise or disturbances. For example even if one would be able to eliminate all stochastic disturbances acting on the system one still has numerical rounding or sensor quantization.
However, when determining observability using the observability matrix, it is completely fine to stop after $k$ terms if you notice that the matrix up till that point is already full rank. Continuing until the $n$th term $C\,A^{n-1}$ is only a nice way to guarantee that the observability is full rank when $(A,C)$ is observable due to the Cayley–Hamilton theorem. I believe that for linear time varying systems, so when the matrices are a function of time $(A(t),C(t))$, it might be that one has to compute beyond the $n$th term in order to determine observability.