Why is this sum the delta function?

Completeness means that any function can be expanded as $$f(x) = \sum_k a_k \phi_k(x),$$ where for brevity I omit spin indices. The expansion coefficients are determined as $$a_k=\int dx \phi_k(x)^*f(x).$$ Combining these two equations we have $$f(x) =\int dx' \sum_k \phi_k(x) \phi_k(x')^*f(x'),$$ from which it follows that $$ \sum_k \phi_k(x')^*\phi_k(x) =\delta(x-x').$$ Often this expression is taken itself as the statement of the completeness.

This is still a property of first quantization. In Dirac notation it becomes quite clear (I find). Define $\psi_k(x)=\langle x|k\rangle$ and $\psi_k(x)^*=\langle k|x\rangle$, dropping spin for now. Here $\psi$ is just a regular function and $|k\rangle$ is the eigenstate with quantum number $k$. For example $\psi_k(x)$ could be the k'th energy state of the harmonic oscillator. Summing these functions over $k$ gives you \begin{align}\sum_k\psi_k(x)\psi_k(x')^*&=\sum_k\langle x|k\rangle\langle k|x'\rangle\\ &=\langle x|\underbrace{\left(\sum_k|k\rangle\langle k|\right)}_1|x'\rangle\\ &=\langle x|x'\rangle\\ &=\delta(x-x') \end{align} Extending to spin starts to get messy but follows the same steps. Define $\psi_k(x)_\alpha=\langle x,\alpha|k\rangle$ where $|x,\alpha\rangle=|x\rangle\otimes|\alpha\rangle$ and $\langle x,\alpha|x',\beta\rangle=\delta_{\alpha,\beta}\delta(x-x')$. Plugging this in gives the desired expression. Why did $\alpha$ and $x$ end up on the bra while $k$ ends up on the ket? Remember that $k$ decribes which state you are looking at while $x$ and $\alpha$ describe which component of that state you are considering.

This is the defining condition for field operators. To show that it is required for a physical theory, we define the interaction density, $I(x)$, whose integral over space is the interaction Hamiltonian. We integrate the Schrodinger equation iteratively and can arrive at the perturbation expansion, or Dyson expansion, which can be written $$U(t) \approx 1+ \sum_{n=1}^\infty \frac{(-1)^n}{n!}\int{d^4x} \cdots \int{d^4x} \mathscr{T} \{I(x_1) \cdots I(x_n) \} $$

Under Lorentz transformation, the order of interactions can be changed in the time-ordered product $\mathscr{T} $ whenever $x_i - x_j$ is space-like. Conversely, Lorentz transformation cannot change the calculation of probabilities under the condition that the initial and final kets are stable states of free particles, as in scattering experiments. The locality condition, or microcausality, follows immediately as a requirement on interaction operators. It states that for any $x,y$ such that $x-y$ is space-like, the commutator of the interaction densities at $x$ and $y$ vanishes, $$[I(y), I(x)] =0 $$ We fulfil this condition by constructing the interaction densities from products of field operators, which obey the same condition for bosons, and the corresponding anti-commutation relation for fermions, which necessarily appear in pairs in the interaction density.

Because we require covariance, vanishing for equal times except at $x=y$ is equivalent to vanishing for spacelike separation. Physically the vanishing of (anti-)commutators for spacelike separations means that effects cannot propagate faster than the speed of light. It can be interpreted as meaning that particles must meet at a point in order to interact, although the point at which they meet has no precise definition, just as the position of a particle has no precise meaning in the general case in quantum mechanics.