A partition problem with order constraints
Intuitively, the intermediate cases should be neither in P, nor NP-hard.  Perhaps it depends exactly on what we mean by "intermediate case".  Here is one interpretation for which we can prove something.

Note: The Exponential-Time Hypothesis, or ETH, is that it is not the case that, for every constant $\epsilon&gt;0$, SAT has an algorithm running in time $2^{n^{\epsilon}}$.  See also this cs.stackexchange discussion.  As far as we know, ETH is true.

Define OP$_c$ to be the restriction of the OrderedPartition problem to instances where $d \le \log^c n$.  Equivalently, to instances where $n \ge 2^{d^{1/c}}$.  Here we intend OP$_c$ to capture what the post means by "intermediate instances".  We show that these instances are not likely to be in P, nor NP-hard.

Lemma 1. If OP$_c$ is in P for all $c$, then ETH fails.

Proof.  Suppose OP$_c$ is in P for all $c$.  That is, for some function $f$, OP$_c$ has an algorithm running in time $n^{f(c)}$. SAT inputs of size $n$ reduce (via Partition as described in the post) to OrderedPartition$[2^{n^{b/c}}, n^b]$, for some constant $b$ and any constant $c&gt;0$.  So, SAT inputs of size $n$ reduce to OP$_c$ instances of size $2^{n^{b/c}}$, which can be solved in time $2^{f(c)n^{b/c}}$ via the algorithm for OP$_c$.  For any $\epsilon&gt;0$, taking, say, $c=2b/\epsilon$, SAT can be solved in time $2^{c' n^{\epsilon/2}} \le 2^{n^{\epsilon}}$ (for large $n$), violating ETH.$~~~~\Box$

Note: It seems likely to me that even OP$_2$ is not in P, but showing something like that would be similar to showing, say, that SAT has no algorithm running in time $2^{\sqrt n}$.

Lemma 2. If OP$_c$ is NP-hard for some $c$, then ETH fails.

Proof.  Suppose OP$_c$ is NP-hard for some $c$.  Then SAT inputs of size $n$ reduce to OP$_c$ in time $O(n^b)$ for some $b$.  That is, to instances of OrdereredPartition$[n^b, d]$ where $d\le \log^c (n^b)$.  As observed in the post such instances can be solved in time 
$n^{O(1)} 2^d = n^{O(1)} 2^{\log^c (n^b)}$, (strongly!) violating ETH.$~~~~\Box$

Probably something cleaner or stronger can be shown.  If I had to guess, I'd define NP$_{d(n)}$ to be the complexity class comprised of those languages that have a non-deterministic poly-time algorithm that, on any input of size $n$, uses at most $d(n)$ non-deterministic guesses.  (Here $d$ could be any function.)  Then OrderedPartition$[n, d(n)]$ is in NP$_{d(n)}$.  Perhaps it is complete for that class under poly-time reductions?  A natural guess for a problem that should be complete for the class would be: given a circuit of size $n$ with $d$ input gates, is there an input that makes the circuit output True?  Or something like that.  (I wonder how this compares to, say, defining SAT$_{p(n)}$ to consist of SAT instances, padded with $p(n)$ useless bits to make the input larger.  When $p(n)$ is super-polynomial but sub-exponential, the problem should be neither NP-hard nor in P.)

p.s. See also Consequences of sub-exponential proofs/algorithms for SAT .