Ruelle-Perron-Frobenius theorem for shift of finite type
The most intuitive explanation I know is the following: suppose that you have a certain amount of mass (I usually picture a pile of sand) that is distributed over $\Sigma_A^+$ according to the density $g\,d\nu$, where $g$ is an arbitrary continuous function and $\nu$ is the eigenmeasure for the transfer operator $\mathcal{L}$. Then you move this mass around according to the shift map: a grain of sand at $\underline{x}$ gets moved to $\sigma(\underline{x})$, for each $\underline{x} \in \Sigma_A^+$. After doing this $n$ times, the sand is distributed according to some new density $g_n \,d\nu$. Then the new density function $g_n$ is given by $g_n = \lambda^{-n} \mathcal{L}^n g$; in other words, the transfer operator tells you how mass moves around (w.r.t. the underlying reference measure $\nu$). This much can be proved just by manipulating the definition of the transfer operator, and is a good exercise if you didn't see it yet. The intuitive content of the limit you marked in green is that as $n\to\infty$, the density functions $g_n$ converge to the eigenfunction $h$, perhaps multiplied by some constant that corresponds to the initial amount of sand; in other words, mass that is spread out according to any initial distribution absolutely continuous w.r.t. $\nu$ will eventually converge towards a distribution with density function given by $h$.
I am not sure if this is really what you are asking for but here is a finite dimensional version of some results of the Perron-Frobenius theorem. I hope this helps you better understand what happens.
Theorem (Perron-Frobenius): If $A\in\Bbb R^{n\times n}$ is a primitive nonnegative matrix, then it has unique (up to scale) left and right eigenvectors $v,h\in\Bbb R^n$ with nonnegative entries. Moreover, $v,h$ have positive entries and their associated eigenvalue is $\rho(A)$, the spectral radius of $A$. In particular, $Ah = \rho(A)h$ and $A^\top v = \rho(A)v$. Furthermore, it holds $$\lim_{k\to \infty} \Big(\frac{A}{\rho(A)}\Big)^{k} =\frac{hv^\top}{\langle v,h\rangle}.$$
Hence, if $\langle v, h\rangle =1 $, then we have $$0=\lim_{k\to \infty} \|\rho(A)^{-k}A^k x-hv^\top x\|=\lim_{k\to \infty} \|\rho(A)^{-k}A^k x-h\langle v,x\rangle\|\qquad \forall x \in \Bbb R^n$$
I don't know what you mean by "intuitively", but this limit in green says that $g\mapsto \nu(g)h$ is the spectral projection associated with the eigenvalue $\lambda$, and iterations of the transfer operator converge to the spectral projection, exactly as in the finite dimensional case as explained by Surb.
Moreover, for shifts of finite type, this convergence in green is exponentially fast, which means that there is a spectral gap : the eigenvalue $\lambda$ is dominant (again exactly as in the finite dimensional case).
To connect all this to the starting point of your question, the measure $dm=hd\nu$ is the (unique) equilibrium state of the function $\phi$ (associated with the dynamical system given by the shift $T$ on the compact space $\Sigma_A⁺$). This can be deduced from the fact that $m$ satisfies a variational principle. This is turn is deduced from the convergence in green, but of course all of this is very well explained in Bowen's book you refer to.