Splay tree with odd number of rotations
It really does not matter for the analysis. The key lemma for analyzing the splay-tree performance is the access lemma. It states that the amortized costs of a splay(x) operation is less than $1+3(r(t)-r(x))$, where $t$ is the root of the tree and $r(u):=\lfloor \log (\text{weight of $u$'s subtree}) \rfloor$.  The weight of a subtree is the sum of the weights of its nodes. (Weights (positive) will be picked depending on the application of the lemma.) The potential function used is $\Phi(T)=\sum_{x \text{ node of } T} r(t)$.

The proof of the access lemma looks at the costs of a single zig/zig-zag/zig-zig etc.  operation. You get 


Costs of a zig or zag  operations are $1+3(r_+(u) - r(u))$, where $r_+$ is the rang after the operation and $u$ is the node rotated upwards.
Costs of zig-zig /zig-zag and symmetric operations are $3(r_+(u) - r(u))$.


If you add these differences up for the operations performed in a single splay(x), you get a telescope sum and what remains is $1+3(r(t)-r(x))$.

If you change the order of the rotations you will get the same sum. The only difference is that now the ''+1'' comes from the first rotation and not from the last rotation. You could even do the zig rotation in the middle. All further (classical) analysis relies on the access lemma.

However, the reason why you perform the single rotation last, is that you do not know if the depth of the node is even or odd in advance.