Let M be a Markov matrix with rows summing to 1.
The interpretation of the left eigenvectors of M is clear. For instance, the first left eigenvector is the stationary distribution of M. And the left second eigenvector represents the distribution of the population after a finite number of steps.
What is the interpretation of the first and second right eigenvectors in the case of a the Markov matrix M?
What are the right eigenvectors of M useful for?
I think it helps to think about it in terms of both sides together. If you have a row-stochastic matrix $P$ corresponding to the Markov process $X_n$, a probability distribution $q$ written as a row vector, and a real-valued function $f$ on the state space written as a column vector, then
$$q P^k f = E[f(X_k)]$$
where $X_0$ is distributed according to $q$. So on the one hand, $q P^k$ is a row vector which describes the evolution of distributions after $k$ steps, while $P^k f$ is a column vector which describes the average evolution of a function after $k$ steps starting from each state. So right eigenvectors are functions $f$ on the state space such that $E[f(X_{k+1})]=\lambda E[f(X_k)]$.
Assuming that $P$ has distinct eigenvalues, is irreducible, and is aperiodic, we wind up with the eigendecomposition
$$q P^k f = q Q \Lambda^k F f$$
where the columns of $Q$ are the left eigenvectors, $\Lambda$ is the eigenvalue matrix, and the rows of $F$ are the right eigenvectors. By an orthogonality argument we wind up with
$$q P^k f = \sum_{i=1}^n \lambda_i^k c_i q_i f_i$$
where $q_i$ are left eigenvectors, $f_i$ are right eigenvectors, and $c_i$ are coefficients which depend on $q$ and $f$. Choose the normalization $f_1=\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$ and isolate out the $i=1$ term:
$$q P^k f = c_1 + \sum_{i=2}^n \lambda_i^k c_i q_i f_i.$$
So $E[f(X_k)]$, for large $k$, is approaching $c_1$, since (under our hypotheses) the other eigenvalues have modulus strictly less than $1$. The main correction term for large $k$ is $c_2 \lambda_2^k q_2 f_2$. So if $q_2 q^T$ and $f_2^T f$ are both significant, then $E[f(X_k)]$ only approaches $c_1$ as fast as $\lambda_2^k$ approaches zero. If $|\lambda_2|$ is nearly $1$, then this convergence is slow.
So roughly speaking, the second right eigenvector $f_2$ is the most metastable function on the state space, while the second left eigenvector $q_2$ is the most metastable perturbation to the equilibrium distribution, in the sense that $q P^k f$ converges most slowly when $f=f_2$ and $q=q_1+\alpha q_2$ for some $\alpha \neq 0$.