Consider a matrix $M \in \mathbb{C}^{2 \times 2}$ and let the eigenvalues of the Gram matrix $M^\dagger M$ (where $\dagger$ denotes the conjugate-transpose) be denoted by $\lambda_1, \lambda_2$. I am interested in knowing whether the condition $0 \le \lambda_k \le 1$ can be understood in terms of some properties of the matrix $M$ itself?
The motivation for this question comes from the fact that $M^\dagger M$ is a Hermitian operator and represents the "observable" in quantum mechanics.
The square root of largest eigenvalue of $M^\dagger M$ is known as the spectral norm of $M$, sometimes denoted $\|M\|$. The eigenvalues of $M^\dagger M$ are necessarily positive, so your condition that the eigenvalues of $M$ are between $0$ and $1$ is equivalent to the statement that $\|M\| \leq 1$.
The spectral norm measures the maximal "growth factor" (relative to the Euclidean norm) associated with the matrix $M$. More concretely, we have $$ \|M\| = \max_{x \neq 0} \frac{\|Mx\|}{\|x\|}, $$ and it generally holds that $\|M x\| \leq \|M\|\cdot \|x\|$ (where $\|Mx\|,\|x\|$ are Euclidean norms of vectors and $\|M\|$ still denotes the spectral norm). The fact that $\|M\| \leq 1$ tells you that for any $x$, we have $\|Mx\| \leq \|x\|$. In other words, $M$ is a contraction.
More generally, the eigenvalues of $M^\dagger M$ are related to the singular value decomposition of $M$.