If the operators $X_i$ can be written as a sum of an annihilation and a creation part$^1$
$$X_i~=~A_i + A^{\dagger}_i, \qquad i~\in ~I, \tag{1}$$
$$\begin{align} A_i|\Omega\rangle~=~0, \qquad & \langle \Omega |A^{\dagger}_i~=~0, \cr
\qquad i~\in~&I,\end{align}\tag{2}$$
where
$$\begin{align} [A_i(t),A_j(t^{\prime})] ~=~& 0, \cr
[A^{\dagger}_i(t),A^{\dagger}_j(t^{\prime})] ~=~& 0, \cr i,j~\in ~&I,\end{align}\tag{3} $$
and
$$\begin{align} [A_i(t),A_j^\dagger(t^{\prime})]
~=~& (c~{\rm number}) \times {\bf 1},\cr i,j~\in~&I,\end{align}\tag{4} $$
i.e. proportional to the identity operator ${\bf 1}$, then one may prove that
$$\begin{align} T(&X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}):\cr
~=~&\langle \Omega | T(X_i(t)X_j(t^{\prime}))|\Omega\rangle ~{\bf 1}.\end{align} \tag{5}$$
Proof of eq. (5): On one hand, the time ordering $T$ is defined as
$$\begin{align} T(&X_i(t)X_j(t^{\prime}))\cr
~=~& \Theta(t-t^{\prime}) X_i(t)X_j(t^{\prime})
+\Theta(t^{\prime}-t) X_j(t^{\prime})X_i(t)\cr
~=~&X_i(t)X_j(t^{\prime}) -\Theta(t^{\prime}-t) [X_i(t),X_j(t^{\prime})]\cr
~\stackrel{(1)+(3)}{=}&~X_i(t)X_j(t^{\prime})\cr &-\Theta(t^{\prime}-t) \left([A_i(t),A^{\dagger}_j(t^{\prime})]+[A^{\dagger}_i(t),A_j(t^{\prime})]\right).\end{align} \tag{6}$$
On the other hand, the normal ordering $::$ moves by definition the creation part to the left of the annihilation part, so that
$$\begin{align}:X_i(t)X_j(t^\prime):~\stackrel{(1)}{=}~& X_i(t)X_j(t^{\prime}) \cr
~-~& [A_i(t),A^{\dagger}_j(t^{\prime})], \end{align}\tag{7}$$
$$ \langle \Omega | :X_i(t)X_j(t^{\prime}):|\Omega\rangle~\stackrel{(1)+(2)}{=}~0.\tag{8}$$
The difference of eqs. (6) and (7) is the lhs. of eq. (5):
$$\begin{align} T(& X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}): \cr
~\stackrel{(6)+(7)}{=}& \Theta(t-t^{\prime})[A_i(t),A^{\dagger}_j(t^{\prime})] \cr
~+~& \Theta(t^{\prime}-t)[A_j(t^{\prime}),A^{\dagger}_i(t)],\end{align}\tag{9}$$
which is proportional to the identity operator ${\bf 1}$ by assumption (4). Now sandwich eq. (9) between the bra $\langle \Omega |$ and the ket $|\Omega\rangle $. Since the rhs. of eq. (9) is proportional to the identity operator ${\bf 1}$, the unsandwiched rhs. must be equal to the sandwiched rhs. times the identity operator ${\bf 1}$. Hence also the unsandwiched lhs. of eq. (9) must also be equal to the sandwiched lhs. times the identity operator ${\bf 1}$. This yields eq. (5). $\Box$
A similar argument applied to eq. (7) yields that
$$\begin{align} X_i(t)&X_j(t^{\prime}) ~-~:X_i(t)X_j(t^{\prime}):\cr
~=~&\langle \Omega | X_i(t)X_j(t^{\prime})|\Omega\rangle ~{\bf 1} \end{align}\tag{10}$$
i.e. a version of eq. (5) without the time-order $T$.
--
$^1$ The operators $A_i$ and $A^{\dagger}_i$ need not be Hermitian conjugates in what follows. We implicitly assume that the vacuum $|\Omega\rangle$ is normalized: $\langle \Omega | \Omega\rangle=1$.
Best Answer
The normal ordening is a way to say: ''we throw away the zero-point energy'' (since it becomes infinity and wa say we only look at energy-differences), or to put it in the words of A. Zee: ''Create before you annihalite''.
The chronological ordening comes in when you calculate the Feynman propagator (also called the Green's function), which is basically the "thing" that you are trying to calculate. If you would look at the Green's function, you know that if $L$ represents your Liouville-equation, for example in the case of the wave-equation:$$L=\frac{\partial^2}{c^2\partial t^2}-\Delta^2$$, the equation for the Green's function $G(r|r')$ hence becomes:$$LG(r|r')=\delta(r-r')$$, which we can use to solve the differential equation represented by $L$ by using a convolution (see the Green's function link).
Now with the Feynman-propagator you do the same thing, and it follows an equivalent equation of the above (only now with some generalized delta-function since we don't werk at equal times)
A book that goes deeper into Wick's theorem is the one of W. Greiner (Field Quantization), which is widely spread on the net. On pages 225-233 Wick's theorem is discussed. After Wick's theorem he starts with QED, this also gives an example of the use of Wick's theorem.
Now I don't know what you hope to find in the answers since the proof is basically an immense ammount of bookkeeping, but basically you make an order-expansion of your product of operators.
Wick-theorem states that:$$T(\hat{A_1}\hat{A_2}\hat{A_3}\hat{A_4}\cdots)=:\hat{A_1}\hat{A_2}\hat{A_3}\hat{A_4}\cdots:+\text{normal ordening with single contractions}+\text{normal ordening with double contractions}+\text{normal ordening with triple contractions}+...$$.
This series goes on untill you run out of stuff to contract (note that you need an even number of operators), so basically op to $n/2$, with $n$ the number of operators.
To use Wick's theorem, one goes order by order. Starting from the zero-th order (which is basically normal-ordening, then going first order (one contraction) and so on. Usually some orders and combinations will simply turn out to be zero (for example when you contract two creation of annihalation operators, or when you contract operators of different fields). You can also exclude terms by using the conservation of momentum and energy (some contractions violate this one).
If you want to compare with Feynman-diagrams, then you need to look at the scattering matrix $S$. This one is developped into a series: $$\hat{S}=\mathbb{I}+\sum\limits_{n=1}^\infty \hat{S}^{(n)}$$, (which is basically the power-series decomposition of your unitary-time evolution operator) for each order, you add a vertex and hence more operators, and thus go to a higher order Feynman-diagram.