Assumptions in basic perturbation theory

Ooofff, there's a lot going on here.

• For one, the choice in the test case in Sakurai's (5.1.5), which is seen more clearly in the equivalent equation (5.1.6), $$ H = begin{pmatrix} E_1^{(0)} & lambda V_{12} lambda V_{21} & E_2^{(0)}end{pmatrix}, tag{5.1.6}$$ in the choice to have $V$ be purely off-diagonal, does not reflect the body of your question. Here the choice to have an off-diagonal $V$ is only speaking about the perturbation itself, and not about the eigenvectors' response to that perturbation; the choice in this limited example has been done simply to keep the exposition as simple as possible and ─ because it is a simple test case ─ it does not have any bearing on the general theory.

• I suppose strictly it doesn't matter whether an additional term is included or not. It just affects the normalisation of the state (which it wouldn't be normalised either way).

  It's a red herring to ask whether the full state is normalized or not. Take the state $$v = N begin{pmatrix} 1 lambda end{pmatrix},$$ normalized to unity by having $N=(1+lambda^2)^{-1/2}$, but taken only to first order in $lambda$: this means that $N approx 1-frac12 lambda^2 approx 1$ doesn't actually kick in, as it only depends quadratically (to leading order) on $lambda$. The full state can indeed be normalized (which may or may not be a good idea ─ it can be easier to account for the normalization elsewhere, i.e. when taking explicit expectation values, and Sakurai devotes an entire section to this) but that only comes in at the $lambda^2$ level or higher. If you're doing second-order perturbation theory then it will matter, but if you're sticking to first-order calculations then it's a moot point.

• As an important point to notice, having diagonal elements for $V$ in the eigenbasis of $H_0$ doesn't do very much, as those elements only change the eigenvalues without actually altering the eigenvectors. This simplicity then feeds into the first-order change in the energy, in (5.1.37), $$ lambda Delta_n^{(1)} = langle n^{(0)} | lambda V | n^{(0)} rangle.tag{5.1.37(a)}$$ Setting that matrix element to zero is equivalent to saying that $Delta_n$ has no first-order dependence, which is obviously not the case in general.

Now, all that said, your intuition that this is related to the state's normalization does have a lot of conceptual depth behind it. The main point where this comes in is at the first full recursive expression for the perturbed eigenvector, eq. (5.1.27), $$ |nrangle = c_n(lambda) |n^{(0)}rangle + frac{1}{E_n^{(0)}-H_0} phi_n (lambda V- Delta_n) |nrangle, tag{5.1.27} $$ where $phi_n = 1- |n^{(0)}ranglelangle n^{(0)}|$ is a projector that excludes the $|n^{(0)}rangle$ direction (and which commutes with $H_0$, so the product $frac{1}{E_n^{(0)}-H_0} phi_n $ also has no $|n^{(0)}rangle$ component), which was introduced to deal with the fact that the shifted original hamiltonian $H_0 - E_n^{(0)}$ is singular at the $|n^{(0)}rangle$ subspace.

The fact that the second term in (5.1.27) has no component along $|n^{(0)}rangle$ is extremely important, because it means that to all orders in $lambda$, the contributions to $|nrangle$ along $|n^{(0)}rangle$ are completely encased within the quasi-normalization function $c_n(lambda)$. Now, as I said, it is perfectly possible to build a perturbation-theory formalism that keeps the full perturbed state normalized to unity, $langle n|nrangle = 1$, but this complicates a number of things, as the various orders of dependence in $c_n(lambda)$ need to be included into the response along the other directions. Instead, Sakurai chooses to normalize his full state to $$ langle n^{0}|nrangle = c_n(lambda) = 1, $$ which does simplify life significantly. (If you want to see how it simplifies life, set up a full Taylor expansion for $c_n(lambda)$ and impose the normalization $langle n|nrangle = 1$, and then see what that does to (5.1.44) and its higher-order analogues.)

And, since you've now completely specified the behaviour of the component of $|nrangle$ along $|n^{(0)}rangle$ for all orders in $lambda$, then you explicitly need to exclude it from the rest of your expansion.