Since the components of the given planar RBM are uncoupled, it suffices to consider a scalar RBM $X$ on the spatial interval $[0,1]$.  Here is a sample path of this process on the temporal interval $[0,10]$.

![sample path][1] 

I will approach local time via the **occupation time**
$$
\int_0^t 1_{[0, \epsilon]}(X(s)) ds
$$
which gives the (random) amount of time the process spends in an $\epsilon$ neighborhood of zero during the interval $[0, t]$. This is a fairly complicated random variable, since it depends on the entire path of the RBM.  However, its expected value is analytically available.  Let
$$
u^{\epsilon}(t,x) = \mathbb{E}_x \int_0^t 1_{[0, \epsilon]}(X(s)) ds \;.
$$ Note that $u^{\epsilon}(t,x)$ satisfies an inhomogeneous initial boundary value problem:
$$
\partial_t u^{\epsilon}(t,x) = 
\frac{1}{2} \partial_{xx} u^{\epsilon}(t,x) + 1_{[0, \epsilon]}(x)  \quad \forall x \in [0,1] \;, \forall t \ge 0 \;,
$$
with initial data $u(0,x)=0$ and pure Neumann boundary conditions $\partial_x u(t,0) = \partial_x u(t,0) = 0$.  By expanding the solution and the inhomogeneity in terms of the [eigenfunctions  of the second derivative operator][2] $\{ e_j(x) \}$ on $[0,1]$ with pure Neumann boundary conditions at $0$ and $1$, one obtains the following explicit solution:
$$
u^{\epsilon}(t,x) = \epsilon t + 2^{3/2} \sum_{j \ge 2}  \left( 1- \exp\left( \frac{t}{2} (j-1)^2 \pi^2  \right) \right) \frac{\sin( (j-1) \pi \epsilon)}{(j-1)^3 \pi^3} e_{j}(x)
$$   
The following figure plots the behavior of $E_x L_t := \lim_{\epsilon \downarrow 0} \frac{u^{\epsilon}(t,x)}{\epsilon}$ as a function of the initial condition of the RBM $X(0)=x$ with $t=1$.  As expected, this quantity decreases with distance from zero. (Since $t=1$, I find it curious that this quantity is greater than unity.)

![local time][3]

To answer the question, just view $x$ in this figure as the vertical component of the cartoon given in the question.  As we already said, what happens in the horizontal component can be treated separately.

**ADD**

By using Monte-Carlo simulation, it is straightforward to obtain even more detail about the random variable $L_t^{\epsilon} = \epsilon^{-1} \int_0^t 1_{[0, \epsilon]}(X(s)) ds$.  Here are graphs of the cumulative distribution function $CDF(s)=\mathbb{P}_x(L_t^{\epsilon} \le s)$ with $t=1$, for the initial conditions  $X(0)=x$ indicated in the figure legend, and for $\epsilon$ sufficiently small.  Logarithmic scaling is adopted to make it easier to compare the CDFs.

![local time cdf][4]


  [1]: https://i.sstatic.net/s06YX.jpg
  [2]: https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative
  [3]: https://i.sstatic.net/T8jnC.jpg
  [4]: https://i.sstatic.net/Ikcbr.jpg