Why do we put two different functions namely $u_{1}(x)$ and $u_{0}(x)$ at places of same function u(x)?
In short, because the method of successive iterations refines the initial guess, which is $u_{0}(x)$, and we have reasons to believe that any initial guess from a wide class of functions will be refined by this process to something that is close to the solution.
Let's consider a simpler problem first. Suppose you are trying to find the limit of a number sequence, defined in the following way:
$$a_{1}=f(x_{0}), \ x_{0}\in \mathbb{R}$$
$$a_{n+1}=f(a_{n})$$
i.e. $a_{n}=\underbrace{f(f(...f(x_{0})))}_{n\text{ times}}$. Suppose $a_{\infty}$ is the limit. Then, if $f(a_{\infty})$ is defined, and the function is continuous in the neighborhood of $a_{\infty}$, $f(a_{\infty})=a_{\infty}$. This means $a_{\infty}$ is a solution of the equation
$$f(x)=x$$
and we can obtain approximations to this solution by choosing the initial guess $x_{0}$ and applying successively $f$ to the previous guess. In particular, if the map $f$ is contracting at some region $D$ of $\mathbb{R}$, i.e. its application reduces the distance between points, $\forall x, y \in D: |f(x)-f(y)|\le \alpha|x-y|$, $\alpha \in [0, 1)$, and $x\in D \Rightarrow f(x) \in D$, then, by a simple "fixed point theorem" a.k.a. theorem on contracting mappings, there exists a unique fixed point of $f$ in $D$, i.e. such point that $f(x)=x$, and the sequence $f(f(...f(x_{0})))$ converges to $x$ for every initial condition $x_{0}\in D$.
The fixed point theorem is actually proved for maps of arbitrary metric spaces, i.e. spaces on which an analogue of distance is defined, not just for mappings $f:\mathbb{R}\to \mathbb{R}$. In general, if you have a set $X$, and a distance function $\rho(x,y): X \times X \to \mathbb{R}_{+}$, satisfying
- $\rho(x,y)=\rho(y,x)$,
- $\rho(x, y)=0 \iff x=y$,
- $\rho(x, y)+\rho(y, z)\ge \rho(x, z)$ (triangle inequality),
then $\rho$ defines the structure of a metric space on $X$, and the fixed point theorem holds: if $f:X \to X$ is such that $\forall x, y\in X: \rho(f(x), f(y))\le \alpha \rho(x, y)$, $\alpha \in [0, 1)$, then $f$ has a unique fixed point, and this fixed point is the limit of successive applications of $f$ to any point $x_0$ of the metric space.
Now, you can define a metric structure on the space of (say, square integrable on a domain $[a, b]$) functions of one real variable $u(x)$, for instance, by taking
$$\rho(u(x), g(x))=\int_{a}^{b}|u(x)-g(x)|^2dx$$
Then one can prove (see, for instance, this answer) that the Volterra operator $A$:
$$A[u(x)](x)=f(x)+\int_{a}^{x}K(x, \xi)u(\xi)d\xi$$
which maps functions to functions, is contracting with respect to the distance functional defined above (or, at any rate, some power of the operator is), if $K(x,y)$ is bounded. This means, by the fixed point theorem, that the limit of successive applications of Volterra operator to any function from $L_{2}[a,b]$ is the same, and is the solution of the equation
$$u(x)=A[u(x)]$$
which is exactly the equation
$$u(x)=f(x)+\int_{a}^{x}K(x, \xi)u(\xi)d\xi$$
you are trying to solve. So you can pick any initial function from the functional space $L_{2}[a,b]$, successively apply the Volterra operator to it, and obtain better and better approximations to the solution of Volterra equation. This answers your second question.
Of course, in practice it is important how fast the function converges, not only the fact that it does, and the choice of initial guess is important, but that is another question.
