Concerning the bit about:
Why do we have the two-particle interaction "sandwiched" between the field operators, but the annihilation/creation operators do not follow the same pattern?
They do follow the same pattern, only when using field operators you are dealing with a continuous variable (i.e. $\textbf{x}$) which makes easier to write the relations in another way. To see this consider for example the kinetic energy operator $T$.
Discrete basis case:
Let $c_k^\dagger, c_k$ are the creation/annihilation operators of momentum eigenstates, and let $a_i^\dagger,a_i$ creation/annihilation operators for a generic (complete) set of states $\{ | i \rangle \}_i$. We have
$$ \tag{D} T =
\sum_{\textbf{k}} \frac{| \textbf{k} |^2}{2m} c_\textbf{k}^\dagger c_\textbf{k} = \sum_{i,j} t_{ij} a_i^\dagger a_j
$$
In both cases you are kind of expanding $T$ in terms of its matrix elements between Fock states. The "kind of" here is important because these are not really expansions in terms of an orthonormal basis for a number of reasons, for example because $a_i^\dagger a_j$ are not projection operators. However, it is still true (as you can readily verify using the (anti)commutation relations) that
$$ \langle \textbf{k} | T | \textbf{k}' \rangle
\equiv \langle 0 | c_\textbf{k} T c_{\textbf{k}'}^\dagger | 0 \rangle
= \delta_{\textbf{k}\textbf{k}'} \frac{ | \textbf{k} |^2}{2m}$$
$$ \langle i | T | j \rangle
\equiv \langle 0 | a_i T a_j^\dagger | 0 \rangle
= t_{ij}$$
Note that the "expanding $T$" interpretation starts to break down when you consider many-particle states, and that in these cases you start appreciating the difference between fermions and bosons. For example for two particle states you have for bosons:
$$ \langle \textbf{k}\textbf{q} | T | \textbf{k}\textbf{q} \rangle
\equiv \langle 0 | c_\textbf{q}c_\textbf{k} T c_\textbf{k}^\dagger c_\textbf{q}^\dagger | 0 \rangle
= \left( \frac{| \textbf{k} |^2}{2m} + \frac{| \textbf{q} |^2}{2m} \right) (1+\delta_{\textbf{k}\textbf{q}})$$
while for fermions:
$$ \langle \textbf{k}\textbf{q} | T | \textbf{k}\textbf{q} \rangle
\equiv \langle 0 | c_\textbf{q}c_\textbf{k} T c_\textbf{k}^\dagger c_\textbf{q}^\dagger | 0 \rangle
= \left( \frac{| \textbf{k} |^2}{2m} + \frac{| \textbf{q} |^2}{2m} \right) (1-\delta_{\textbf{k}\textbf{q}})$$
Continuous basis case
Using the field operators $\psi(\textbf{x})$ and $\psi^\dagger(\textbf{x})$ we write $T$ as:
$$ \tag{C} T = \int d^3 x \psi^\dagger(\textbf{x}) \left (\frac{-1}{2m} \nabla^2 \right) \psi(\textbf{x}) $$
so how is this similar to something like (D)?
To better see this lets assume we are in a 1D space (so $\textbf{x} \approx x_n$, $\psi(\textbf{x}) \approx \psi_n $) and switch to discrete varibles. We then have
$$ \nabla^2 \psi(\textbf{x}) \approx ( \psi_{n+2} - 2 \psi_{n+1} + \psi_n ) $$
and using this $T$ becomes
$$ \tag{R} T = \frac{-1}{2m} \sum_n \psi_n^\dagger ( \psi_{n+2} - 2 \psi_{n+1} + \psi_n ) = \sum_{n,m} \psi^\dagger_n T_{nm} \psi_m = \sum_{n,m} T_{nm} \psi^\dagger_n \psi_m $$
where we have defined
$$ T_{nm} \equiv \frac{-1}{2m} ( \delta_{n,m+2} - 2 \delta_{n,m+1} + \delta_{n,m} ) $$
as you can see (R) is again a form like (D), and the kinetic operator is no longer "sandwiched" between field operators (but it has acquired a more cumbersome form).
You're final equation is false. It is not true that,
$$
\left\langle ... n _i ... \right| a = \sqrt{ n _i } \left\langle ... n _i - 1 ... \right|
$$
The correct result is
$$
\left\langle ... n _i ... \right| a = \left\langle ... n _i + 1 ... \right| \sqrt{ n _i +1}
$$
as you have just proven.
A non-Hermitian operator doesn't act in the same way to the left as it does to the right. The rules of working with bra's and ket's are if you are acting backwards with the operator then you need to use the Hermitian adjoint (see for example Wikipedia). Otherwise you end up with inconsistencies as you did above. More formally, for any operator $ A$ and states $ \left| \alpha \right\rangle $ and $ \left| \beta \right\rangle $ you can write,
\begin{equation}
\left\langle \alpha \right| A \left| \beta \right\rangle \equiv \left\langle \alpha |A \beta \right\rangle = \langle A ^\dagger \alpha | \beta \rangle
\end{equation}
and only for a Hermitian operator can you omit the dagger.
Best Answer
Creation and annihilation operators are ladder operator in the sense that they raise and lower respectively the quantum numbers of a state (such as e.g. the number of particles in an harmonic oscilattor, the angular momentum for spins, etc...). If they were hermitian, that is $a=a^\dagger$, the same operator $a$ should lower and raise the quantum number at the same time spoiling its very definition. Perhaps, you can think of the ``number'' operator $a^\dagger a$ as a sort of hermitian cousin of $a$ or $a^\dagger$, and indeed it does not raise or lower the quantum numbers but it simply counts.