Well, first of, the dummy variable is interpreted as a change in intercept. That is, your coefficient $\beta_3$ gives you the difference in the intercept when $D=1$, i.e. when $D=1$, the intercept is $\beta_0 + \beta_3$. That interpretation doesn't change when adding the squared $x_1$.
Now, the point of adding a squared to the series is that you assume that the relationship wears off at a certain point. Looking at your second equation
$$y = \beta _0 + \beta_1x_1+\beta_2x_1^2+\beta_3 D + \varepsilon$$
Taking the derivate w.r.t. $x_1$ yields
$$\frac{\delta y}{\delta x_1} = \beta_1 + 2\beta_2 x_1$$
Solving this equation gives you the turning point of the relationship. As user1493368 explained, this is indeed reflecting an inverse U-shape if $\beta_1<0$ and vice versa. Take the following example:
$$\hat{y} = 1.3 + 0.42 x_1 - 0.32 x_1^2 + 0.14D$$
The derivative w.r.t. $x_1$ is
$$\frac{\delta y}{\delta x_1} = 0.42 - 2\times 0.32 x_1 $$
Solving for $x_1$ gives you
$$\frac{\delta y}{\delta x_1} = 0 \iff x_1 \approx 0.66 $$
That is the point at which the relationship has its turning point. You can take a look at Wolfram-Alpha's output for the above function, for some visualization of your problem.
Remember, when interpreting the ceteris paribus effect of a change in $x_1$ on $y$, you have to look at the equation:
$$\Delta y = (\beta_1 + 2\beta_2x_1)\Delta x$$
That is, you can not interpret $\beta_1$ in isolation, once you added the squared regressor $x_1^2$!
Regarding your insignificant $D$ after including the squared $x_1$, it points towards misspecification bias.
