I admitted for the current stiff PDE education system, it is hard to find the examples of solving PDEs with sightly innovative types of conditions.
So it is the time to use the brains ourselves.
For your first example, it gives the message that the roles of $x$ and $t$ can be interchanged.
Please use the general solution of the form $u(x,t)=f\left(t+\dfrac{x}{2}\right)+g\left(t-\dfrac{x}{2}\right)+4t^2-x^2$ instead for convenience.
$u(0,t)=t$ :
$f(t)+g(t)+4t^2=t$
$f(t)+g(t)=t-4t^2~......(1)$
$u_x(x,t)=f_x\left(t+\dfrac{x}{2}\right)+g_x\left(t-\dfrac{x}{2}\right)-2x=\dfrac{1}{2}f_t\left(t+\dfrac{x}{2}\right)-\dfrac{1}{2}g_t\left(t-\dfrac{x}{2}\right)-2x$
$u_x(0,t)=0$ :
$\dfrac{1}{2}f_t(t)-\dfrac{1}{2}g_t(t)=0$
$f_t(t)-g_t(t)=0$
$f(t)-g(t)=c~......(2)$
$\therefore f(t)=\dfrac{t}{2}-2t^2+\dfrac{c}{2},g(t)=\dfrac{t}{2}-2t^2-\dfrac{c}{2}$
$\therefore u(x,t)=\dfrac{t}{2}+\dfrac{x}{4}-2\left(t+\dfrac{x}{2}\right)^2+\dfrac{c}{2}+\dfrac{t}{2}-\dfrac{x}{4}-2\left(t-\dfrac{x}{2}\right)^2-\dfrac{c}{2}+4t^2-x^2=t-2x^2$
For your second example,
Please use the general solution of the form $u(x,t)=f(2t+x)+g(2t-x)+\dfrac{t^2}{2}-\dfrac{x^2}{8}$ instead for convenience.
$u(x,x)=x^2$ :
$f(3x)+g(x)+\dfrac{x^2}{2}-\dfrac{x^2}{8}=x^2$
$f(3x)+g(x)=\dfrac{5x^2}{8}~......(1)$
$u_t(x,t)=f_t(2t+x)+g_t(2t-x)+t=2f_x(2t+x)+2g_x(2t-x)+t$
$u_t(x,x)=x$ :
$2f_x(3x)+2g_x(x)+x=x$
$3f_x(3x)+3g_x(x)=0$
$f(3x)+3g(x)=c~......(2)$
$\therefore f(x)=\dfrac{5x^2}{48}-\dfrac{c}{2},g(x)=\dfrac{c}{2}-\dfrac{5x^2}{16}$
$\therefore u(x,t)=\dfrac{5(2t+x)^2}{48}-\dfrac{c}{2}+\dfrac{c}{2}-\dfrac{5(2t-x)^2}{16}+\dfrac{t^2}{2}-\dfrac{x^2}{8}=-\dfrac{x^2-5xt+t^2}{3}$
Your two examples are having the conditions in same positions. Note that for the cases of having the conditions in different positions, their PDEs may still have infinitely many solutions.
For example $u_{tt}=u_{xx}$ with $u(x,0)=0$ and $u(0,t)=0$ ,
The general solution is $u(x,t)=f(x+t)+g(x-t)$
$u(x,0)=0$ :
$f(x)+g(x)=0~......(1)$
$u(0,t)=0$ :
$f(t)+g(-t)=0~......(2)$
$\therefore f(t)-f(-t)=0$
$f(t)=\Phi(t)$ , where $\Phi(t)$ is any even function of $t$
$\therefore g(x)=-\Phi(x)$ , where $\Phi(x)$ is any even function of $x$
$\therefore u(x,t)=\Phi(x+t)-\Phi(x-t)$ , where $\Phi(x)$ is any even function of $x$
Another good example is Does the wave equation require an initial function for one of its derivative?.
Your interpretation is correct. The principal at work here is the method of images. The idea is that the "reflecting" constraint $v(0,t)$ can be eliminated by expanding the domain and adding an initial condition which automatically enforces the constraint. The idea is that by using the odd extension for the auxiliary PDE $u$, you introduce the symmetry $u(x,t)=-u(-x,t)$, which is preserved by the evolution of the PDE. This in particular means that $u(0,t)=0$ without having to impose any constants. You can then "forget" the $x<0$ half of the domain and the inteference from the virtual half can be reinterpreted as the reflection off of the barrier at $x=0$.
For a visualization, notice that the centerpoint of the above animation does not move; by covering up the right half, it will instead look like the right pulse is reflecting off this centerpoint.
Best Answer
This is not an answer, but an unsuccessful attempt to reconcile the intuition of d'Alembert's formula with the explanation you've given, as well as an iteration of a cry for help since in my humble opinion there are same gaps in the intuition, which your question calls upon.
The general solution is $$2u(x,t)=f(x-t)+f(x+t)+\int_{x-t}^{x+t}g(\tau)d\tau.$$ As you pointed out, if $g\equiv0$ then the initial wave $f$ splits in two and the solution goes in opposite directions (I imagine a string stretched and let go from rest). This case makes sense.
For the other situation, if $f\equiv0$, then for any $x$, $u(x,t)$ approaches the integral of $g$ as $t$ goes to infinity. This makes no intuitive sense, since straight from the formula we know then that if say, $g$ is any bump function with area 1, then for any $x$, $u(x,t)$ is eventually $1$, which is like saying "if you pluck a string with unit velocity, the entire string will eventually sit one unit in the direction that you've plucked it". This statement of course makes no sense physically...