From the well-known definition of Feynman propagator that you have put above, we can re-write it as
\begin{equation}
D_F (x-y)=\begin{cases}
\langle 0|\phi(x), \phi(y)| 0 \rangle,\; x^0>y^0 \\
\langle 0|\phi(y), \phi(x) | 0\rangle,\; y^0>x^0
\end{cases}.\tag{1}
\end{equation}
Let's decomposed the scalar field to a creation and anihilation parts, respectively
$$\phi(x)=\phi_+(x)+\phi_-(x)=\frac 1 {\sqrt{(2\pi)^3}}\int \frac{d^3k}{\sqrt{2 \omega_k}}\bigg( \hat a^{\dagger}_{\vec k} e^{-ik\cdot x} +\hat a_{\vec k} e^{ik\cdot x} \bigg)$$
(where $k \cdot x = -k_0 x^0+ \vec k \cdot \vec x$)
Therefore we have
$$\phi(x)\phi(y)=\phi_+(x)\phi_+(y)+\phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y)+\phi_-(x)\phi_-(y)$$
and then
$$\boxed{\langle 0|\phi(x) \phi(y)| 0 \rangle=\langle 0|\phi_-(x) \phi_+(y)| 0 \rangle}\tag{2}$$
Since
\begin{eqnarray}
\langle 0| \phi_+ (x) &\sim& \langle 0| \hat a^\dagger_{\vec k}=0\;,\\
\phi_-(y)|0 \rangle &\sim& \hat a_{\vec k}| 0 \rangle =0 \;.
\end{eqnarray}
Similarly, we have
$$\boxed{\langle 0|\phi(y) \phi(x)| 0 \rangle=\langle 0|\phi_-(y), \phi_+(x)| 0 \rangle} \tag{3}$$
The important trick is an observation that
\begin{eqnarray}
\langle 0|\phi_-(x) \phi_+(y)| 0 \rangle &=& [\phi_-(x), \phi_+(y)] \;\;\; \big(\equiv i \Delta^{(+)}(x-y)\big)\;,\\
\langle 0|\phi_-(y) \phi_+(x)| 0 \rangle &=& [\phi_-(y), \phi_+(x)] \;\;\; \big(\equiv i \Delta^{(+)}(y-x)\big)\;\tag{4}
\end{eqnarray}
These follow from the properties of creation and anihilation operators
$$\boxed{{ \langle 0| \hat a_{\vec k}\hat a^\dagger_{\vec k'} |0\rangle=\langle \vec k| \vec k' \rangle =\delta^{(3)}(\vec k- \vec k')=[\hat a_{\vec k},\hat a^\dagger_{\vec k'}] }}$$
So if as you said $D_F (x-y)=\overline{\phi(x)\phi(y)}$, from (1),(2),(3), and (4) the correct definition of the Wick contraction would be
$$
\overline{\phi(x) \phi(y)}=
\begin{cases}
[\phi_-(x), \phi_+(y)], x^0>y^0 \\
[\phi_-(y), \phi_+(x)], y^0>x^0
\end{cases}.\tag{5}
$$
or $\overline{\phi(x) \phi(y)}=\theta(x^0-y^0)i\Delta^{(+)}(x-y)+\theta(y^0-x^0)i\Delta^{(+)}(y-x)\;,$ where the step function is defined by
$$
\theta(x^0-y^0)=\begin{cases}
1,\; x^0>y^0 \\
0,\; y^0>x^0.
\end{cases}
$$
By these details we can proof your last equation as the following
\begin{eqnarray}
T \phi(x)\phi(y) &=& \theta(x^0-y^0)\phi(x)\phi(y) +\theta(y^0-x^0)\phi(y)\phi(x)\;,\\
&=&\theta(x^0-y^0)\big[ \phi_+(x)\phi_+(y)+\phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y)+\phi_-(x)\phi_-(y) \big]\\
&&+\theta(y^0-x^0)\big[ \phi_+(y)\phi_+(x)+\phi_+(y)\phi_-(x)+\phi_-(y)\phi_+(x)+\phi_-(y)\phi_-(x) \big]\;.\tag 6
\end{eqnarray}
Then we will make used of $$[\phi_+(x),\phi_+(y)]=[\phi_-(x),\phi_-(y)]=0$$
or $$\phi_+(x)\phi_+(y)=\phi_+(y)\phi_+(x)$$ $$\phi_-(x)\phi_-(y)=\phi_-(y)\phi_-(x)\tag 7 $$
in addition $$\boxed{\theta(x^0-y^0)+\theta(y^0-x^0)=1} \tag 8$$
Back to (6), with these tools, we then have
\begin{eqnarray}
T\phi(x)\phi(y) &=& \phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)\\
&&+\theta(x^0-y^0)\big[ \phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y) \big]\\
&&+\theta(y^0-x^0)\big[ \phi_+(y)\phi_-(x)+\phi_-(y)\phi_+(x) \big]\;,\\
&=& \phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)\\
&&+\theta(x^0-y^0)\big[ \boxed{\boxed{ \phi_+(x)\phi_-(y)}}+\phi_-(x)\phi_+(y) -\phi_+(y) \phi_-(x)+\boxed{\phi_+(y) \phi_-(x)}\big]\\
&&+\theta(y^0-x^0)\big[ \boxed{ \phi_+(y)\phi_-(x)}+\phi_-(y)\phi_+(x)-\phi_+(x) \phi_-(y)+\boxed{\boxed{ \phi_+(x) \phi_-(y)}}\big]\;,\\
&=&\phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)+ \boxed{\boxed{ \phi_+(x)\phi_-(y)}}+ \boxed{\phi_+(y)\phi_-(x)}\\
&&+\theta(x^0-y^0)[\phi_-(x),\phi_+(y) ]\\
&&+\theta(y^0-x^0)[\phi_-(y),\phi_+(x) ]\;,\\
&=& :\phi(x)\phi(y):+ \overline{\phi(x)\phi(y)}\;.
\end{eqnarray}
As we want.
