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.