$\left(1\right)\because A\left(-2,2\right)$，$B\left(-8,-2\right)$，将线段$AB$平移得到线段$DC$，点$A$的对应点$D$在$x$轴上，点$B$的对应点$C$在$y$轴上，
$\therefore$点$A$的纵坐标减$2$，点$B$的横坐标加$8$，
$\therefore$点$A$的对应点$D$的坐标是$\left(-2+8,2-2\right)$，即$D\left(6,0\right)$，
点$B$的对应点$C$的坐标是$\left(-8+8,-2-2\right)$，即$C\left(0,-4\right)$；
$(2)$如图，设$AD$交$y$轴于点$Q$，连接$OA$，设$P\left(0,y\right)$，$Q\left(0,t\right)$，

$\because A\left(-2,2\right)$，$C\left(0,-4\right)$，$D\left(6,0\right)$，
$\therefore OQ=t$，$OD=6$，$CP=|-4-y|$，
$\because S_{\triangle AOQ}+S_{\triangle DOQ}=S_{\triangle AOD}$，
$\therefore \frac{1}{2}\times 2t+\frac{1}{2}\times 6t=\frac{1}{2}\times 6\times 2$，
解得：$t=\frac{3}{2}$，
$\therefore Q(0$，$\frac{3}{2})$，
$\therefore PQ=\frac{3}{2}-y$，
$\because S_{\triangle APD}=2S_{\triangle CPD}$，即$S_{\triangle APQ}+S_{\triangle DPQ}=2S_{\triangle CPD}$，
$\therefore \frac{1}{2}\times (\frac{3}{2}-y)\times 2+\frac{1}{2}\times (\frac{3}{2}-y)\times 6=2\times \frac{1}{2}\times |-4-y|\times 6$，
解得：$y=-\frac{9}{5}$或$-15$，
$\therefore$点$P$的坐标为$(0$，$-\frac{9}{5})$或$\left(0,-15\right)$；
$(3)$①$S_{\triangle DEM}=S_{\triangle CFM}$，理由如下：如图，连接$OM$，设$M\left(x,y\right)$，

$\because$动点$E$从点$D$出发向左运动，同时动点$F$从点$C$出发向上运动，两个点的运动速度之比为$3:2$，
$\therefore E\left(6-3m,0\right)$，$F\left(0,-4+2m\right)$，
$\because$直线$DF$和$CE$的交点$M$在第二象限，
$\therefore 6-3m \lt 0$，
$\therefore m \gt 2$，
则$DE=3m$，$CF=2m$，
又$\because C\left(0,-4\right)$，$D\left(6,0\right)$，
$\therefore OC=4$，$OD=6$，
$\because S_{\triangle DOM}=S_{\triangle DOF}+S_{\triangle FOM}$，
$\therefore \frac{1}{2}OD\cdot y=\frac{1}{2}OD\cdot OF+\frac{1}{2}OF\cdot \left(-x\right)$，
即$\frac{1}{2}\times 6\cdot y=\frac{1}{2}\times 6\cdot \left(2m-4\right)+\frac{1}{2}(2m-4)\cdot \left(-x\right)$①，
同理$\frac{1}{2}\times 4\cdot \left(-x\right)=\frac{1}{2}\times 4\cdot \left(3m-6\right)+\frac{1}{2}(3m-6)\cdot y_{②}$，
联立①②，得$\left\{\begin{array}{l}{y=\frac{4}{6-3m}x-4}\\{y=\frac{2-m}{3}x+2m-4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=\frac{6m-12}{m-4}}\\{y=\frac{4m-8}{4-m}}\end{array}\right.$，
$\therefore M(\frac{6m-12}{m-4}$，$\frac{4m-8}{4-m})$，
$\therefore S_{\triangle DEM}=\frac{1}{2}\times 3m\times \frac{4m-8}{4-m}=\frac{6{m}^{2}-12m}{4-m}$，$S_{\triangle CFM}=\frac{1}{2}\times 2m\times \frac{12-6m}{m-4}=\frac{6{m}^{2}-12m}{4-m}$，
$\therefore S_{\triangle DEM}=S_{\triangle CFM}$；
②当点$M$在第二象限时，如图，

由①得：$E\left(6-3m,0\right)$，$F\left(0,-4+2m\right)$，$M(\frac{6m-12}{m-4}$，$\frac{4m-8}{4-m})$，$S_{\triangle DEM}=\frac{6{m}^{2}-12m}{4-m}$，
$\because S_{\triangle CDE}=\frac{1}{2}\times 3m\times 4=6m$，
$\therefore S_{\triangle CDM}=S_{\triangle CDE}+S_{\triangle MDE}=6m+\frac{6{m}^{2}-12m}{4-m}=\frac{12m}{4-m}$，
$\because S_{\triangle CDM}=14$，
$\therefore \frac{12m}{4-m}=14$，
解得：$m=\frac{28}{13}$，
当$m=\frac{28}{13}$时，$\frac{6m-12}{m-4}=\frac{6×\frac{28}{13}-12}{\frac{28}{13}-4}=-\frac{1}{2}$，$\frac{4m-8}{4-m}=\frac{4×\frac{28}{13}-8}{4-\frac{28}{13}}=\frac{1}{3}$，
$\therefore$点$M$的坐标为$(-\frac{1}{2}$，$\frac{1}{3})$；
当点$M$在第四象限时，如图，

则$E\left(6+3n,0\right)$，$F\left(0,-4-2n\right)$，
同①可得：$M(\frac{6n+12}{n+4}$，$-\frac{4n+8}{n+4})$，
$\therefore S_{\triangle DEM}=\frac{1}{2}\times 3n\times \frac{4n+8}{n+4}=\frac{6{n}^{2}+12n}{n+4}$，$S_{\triangle CDE}=\frac{1}{2}\times 3n\times 4=6n$，
$\therefore S_{\triangle CDM}=S_{\triangle CDE}-S_{\triangle DEM}=6n-\frac{6{n}^{2}+12n}{n+4}=\frac{12n}{n+4}$，
$\because S_{\triangle CDM}=14$，
$\therefore \frac{12n}{n+4}=14$，
解得：$n=-28$，
$\therefore M(\frac{13}{2}$，$-\frac{13}{3})$；
综上所述，点$M$的坐标为$(-\frac{1}{2}$，$\frac{1}{3}$）或（$\frac{13}{2}$，$-\frac{13}{3})$.

$\left(1\right)\because A\left(-2,2\right)$，$B\left(-8,-2\right)$，将线段$AB$平移得到线段$DC$，点$A$的对应点$D$在$x$轴上，点$B$的对应点$C$在$y$轴上，
$\therefore$点$A$的纵坐标减$2$，点$B$的横坐标加$8$，
$\therefore$点$A$的对应点$D$的坐标是$\left(-2+8,2-2\right)$，即$D\left(6,0\right)$，
点$B$的对应点$C$的坐标是$\left(-8+8,-2-2\right)$，即$C\left(0,-4\right)$；
$(2)$如图，设$AD$交$y$轴于点$Q$，连接$OA$，设$P\left(0,y\right)$，$Q\left(0,t\right)$，

$\because A\left(-2,2\right)$，$C\left(0,-4\right)$，$D\left(6,0\right)$，
$\therefore OQ=t$，$OD=6$，$CP=|-4-y|$，
$\because S_{\triangle AOQ}+S_{\triangle DOQ}=S_{\triangle AOD}$，
$\therefore \frac{1}{2}\times 2t+\frac{1}{2}\times 6t=\frac{1}{2}\times 6\times 2$，
解得：$t=\frac{3}{2}$，
$\therefore Q(0$，$\frac{3}{2})$，
$\therefore PQ=\frac{3}{2}-y$，
$\because S_{\triangle APD}=2S_{\triangle CPD}$，即$S_{\triangle APQ}+S_{\triangle DPQ}=2S_{\triangle CPD}$，
$\therefore \frac{1}{2}\times (\frac{3}{2}-y)\times 2+\frac{1}{2}\times (\frac{3}{2}-y)\times 6=2\times \frac{1}{2}\times |-4-y|\times 6$，
解得：$y=-\frac{9}{5}$或$-15$，
$\therefore$点$P$的坐标为$(0$，$-\frac{9}{5})$或$\left(0,-15\right)$；
$(3)$①$S_{\triangle DEM}=S_{\triangle CFM}$，理由如下：如图，连接$OM$，设$M\left(x,y\right)$，

$\because$动点$E$从点$D$出发向左运动，同时动点$F$从点$C$出发向上运动，两个点的运动速度之比为$3:2$，
$\therefore E\left(6-3m,0\right)$，$F\left(0,-4+2m\right)$，
$\because$直线$DF$和$CE$的交点$M$在第二象限，
$\therefore 6-3m \lt 0$，
$\therefore m \gt 2$，
则$DE=3m$，$CF=2m$，
又$\because C\left(0,-4\right)$，$D\left(6,0\right)$，
$\therefore OC=4$，$OD=6$，
$\because S_{\triangle DOM}=S_{\triangle DOF}+S_{\triangle FOM}$，
$\therefore \frac{1}{2}OD\cdot y=\frac{1}{2}OD\cdot OF+\frac{1}{2}OF\cdot \left(-x\right)$，
即$\frac{1}{2}\times 6\cdot y=\frac{1}{2}\times 6\cdot \left(2m-4\right)+\frac{1}{2}(2m-4)\cdot \left(-x\right)$①，
同理$\frac{1}{2}\times 4\cdot \left(-x\right)=\frac{1}{2}\times 4\cdot \left(3m-6\right)+\frac{1}{2}(3m-6)\cdot y_{②}$，
联立①②，得$\left\{\begin{array}{l}{y=\frac{4}{6-3m}x-4}\\{y=\frac{2-m}{3}x+2m-4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=\frac{6m-12}{m-4}}\\{y=\frac{4m-8}{4-m}}\end{array}\right.$，
$\therefore M(\frac{6m-12}{m-4}$，$\frac{4m-8}{4-m})$，
$\therefore S_{\triangle DEM}=\frac{1}{2}\times 3m\times \frac{4m-8}{4-m}=\frac{6{m}^{2}-12m}{4-m}$，$S_{\triangle CFM}=\frac{1}{2}\times 2m\times \frac{12-6m}{m-4}=\frac{6{m}^{2}-12m}{4-m}$，
$\therefore S_{\triangle DEM}=S_{\triangle CFM}$；
②当点$M$在第二象限时，如图，

由①得：$E\left(6-3m,0\right)$，$F\left(0,-4+2m\right)$，$M(\frac{6m-12}{m-4}$，$\frac{4m-8}{4-m})$，$S_{\triangle DEM}=\frac{6{m}^{2}-12m}{4-m}$，
$\because S_{\triangle CDE}=\frac{1}{2}\times 3m\times 4=6m$，
$\therefore S_{\triangle CDM}=S_{\triangle CDE}+S_{\triangle MDE}=6m+\frac{6{m}^{2}-12m}{4-m}=\frac{12m}{4-m}$，
$\because S_{\triangle CDM}=14$，
$\therefore \frac{12m}{4-m}=14$，
解得：$m=\frac{28}{13}$，
当$m=\frac{28}{13}$时，$\frac{6m-12}{m-4}=\frac{6×\frac{28}{13}-12}{\frac{28}{13}-4}=-\frac{1}{2}$，$\frac{4m-8}{4-m}=\frac{4×\frac{28}{13}-8}{4-\frac{28}{13}}=\frac{1}{3}$，
$\therefore$点$M$的坐标为$(-\frac{1}{2}$，$\frac{1}{3})$；
当点$M$在第四象限时，如图，

则$E\left(6+3n,0\right)$，$F\left(0,-4-2n\right)$，
同①可得：$M(\frac{6n+12}{n+4}$，$-\frac{4n+8}{n+4})$，
$\therefore S_{\triangle DEM}=\frac{1}{2}\times 3n\times \frac{4n+8}{n+4}=\frac{6{n}^{2}+12n}{n+4}$，$S_{\triangle CDE}=\frac{1}{2}\times 3n\times 4=6n$，
$\therefore S_{\triangle CDM}=S_{\triangle CDE}-S_{\triangle DEM}=6n-\frac{6{n}^{2}+12n}{n+4}=\frac{12n}{n+4}$，
$\because S_{\triangle CDM}=14$，
$\therefore \frac{12n}{n+4}=14$，
解得：$n=-28$，
$\therefore M(\frac{13}{2}$，$-\frac{13}{3})$；
综上所述，点$M$的坐标为$(-\frac{1}{2}$，$\frac{1}{3}$）或（$\frac{13}{2}$，$-\frac{13}{3})$.