Is there a clear and intuitive meaning to the eigenvectors and eigenvalues of a density matrix?

In general, the density matrix of a given system can always be written in the form $$ \rho = \sum_i p_i |\phi_i\rangle\langle\phi_i|, \tag 1 $$ representing among other things a probabilistic mixture in which the pure state $|\phi_i\rangle$ is prepared with probability $p_i$, but this decomposition is generally not unique. The clearest example of this is the maximally mixed state on, say, a two-level system with orthonormal basis $\{|0⟩,|1⟩\}$, $$ \rho = \frac12\bigg[|0⟩⟨0|+|1⟩⟨1|\bigg], $$ which has exactly the same form on any orthonormal basis for the space.

Generally speaking, though, the eigenvalues and eigenvectors of a given density matrix $\rho$ provide a set of states and weights such that $\rho$ can be written as in $(1)$ - but with the added guarantee that the $|\phi_i⟩$ are orthogonal.

This does not uniquely specify the states in question, because if any eigenvalue $p_i$ is degenerate then there will be a two-dimensional (or bigger) subspace within which any orthonormal basis is equally valid, but that kind of undefinedness is just an intrinsic part of the structure.

The density operator is defined as

$$\rho = \sum_{i=1}^N p_i |X_i\rangle\langle X_i|$$

for a set of $N$ states $|X_i\rangle$ that occur with probability $p_i$. These states form an orthonormal basis to ensure the normalization condition $Tr(\rho)=1$.

We can see from this definition that the eigenvalues are the probabilities $p_i$ that are real numbers; the operator assigns simply the probability to some pure state (that is not in a superposition with other states) without changing this state. Therefore, we know from linear algebra that if eigenvalues are real, then the density matrix must be hermitean.

Another way to see hermiticity is to check $\langle b|\rho = (\rho|b\rangle)^*$ for any state $|b\rangle$.

From hermiticity follows that eigenvalues and eigenvectors exists (spectral theorem).