The position of the dog relative to the human is a Markov chain, which is symmetric by inspection (there are three cases to consider: interior squares, "edge squares," and "corner squares"). This Markov chain splits up into two irreducible ergodic components corresponding to even and odd squares. Finally, any symmetric, irreducible, ergodic Markov chain has uniform stationary distribution. Since the number of squares at distance $d$ is $4d$ (except for $d = 0$), it follows that the distribution of the distances will be linear (again except for $d = 0$).

The position of the dog relative to the human is a Markov chain, which is symmetric by inspection (there are three cases to consider: interior squares, "edge squares," and "corner squares"). This Markov chain splits up into two irreducible ergodic components corresponding to even and odd squares. Finally, any symmetric, irreducible, ergodic Markov chain has uniform stationary distribution. Since the number of squares at distance $d$ is $4d$ (except for $d = 0$), it follows that the distribution of the distances will be linear (again except for $d = 0$).

Here's how to show symmetry. If you're in an interior square, you move in each cardinal direction with probability $\frac{1}{16}$ and each diagonal direction with probability $\frac{2}{16}$. If you're in an edge square, you move diagonally off the edge with probability $\frac{2}{16}$, cardinally off the edge in each of two directions with probability $\frac{1}{16}$, and along the edge in each direction with probability $\frac{3}{16}$. If you're in a corner square, you move diagonally to an edge in each of two directions with probability $\frac{3}{16}$, and into the interior in one cardinal direction with probability $\frac{1}{16}$. We then observe that the probabilities of moving to each square from each other square are equal to the probabilities of moving the other way. (e.g., if we're on an edge next to a corner, we move into the corner with probability $\frac{3}{16}$ and then move out to that same edge square with the same probability.)

The position of the dog relative to the human is a Markov chain, so the distribution of the dog's position can be computed directly for any given $\lambda$. Computing this for various values of $\lambda$, I get that the dog is uniformly distributed on either the odd or the even squares, depending on where it starts, and hence the time spent at each distance will be linear in the distance (with an anomaly at distance 0 because there's an extra square there).

Of course, this is not a proof (at least not one that works for all $\lambda$), but perhaps it can be shown directly that the Markov chain will always be uniform.

The position of the dog relative to the human is a Markov chain, which is symmetric by inspection (there are three cases to consider: interior squares, "edge squares," and "corner squares"). This Markov chain splits up into two irreducible ergodic components corresponding to even and odd squares. Finally, any symmetric, irreducible, ergodic Markov chain has uniform stationary distribution. Since the number of squares at distance $d$ is $4d$ (except for $d = 0$), it follows that the distribution of the distances will be linear (again except for $d = 0$).