This second answer specifically addresses the issue of trying to interpret $v\circ x$ when $x$ is not regular and $v$ is a distribution. I will tie this into the edit to illustrate the problem.
The question is, give that $x$ is the function that is $e^t$ when $t < 0$ and $2e^t$ when $t > 0$, what is the distribution $\delta_1\circ x$. Specifically, is it the case that
$$ \int \delta_1(x(t)) \omega(t) = \omega(0) $$
for all test functions $\omega$?
I claim that this cannot be a well-defined operation. One way to see this is through approximations to the identity. Ideally we would like it to be the case that whatever the definition is, if we take a sequence of smooth functions $f_\epsilon$ that converges (as a distribution) to $\delta_1$ as $\epsilon \to 0$, that
$$ \int f_\epsilon(x(t)) \omega(t) \to \omega(0) $$
Case 1: symmetric bumps
Let $f$ be a smooth, even, bump function (here's a concrete example). Let
$$ f_\epsilon(x) = \frac1\epsilon f(\frac{x-1}{\epsilon}) $$
Standard arguments (see the above linked Wikipedia page) shows that for any test function $g$, you have
$$ \int f_\epsilon(x) g(x) dx \to g(1) $$
Now look at $f_\epsilon(x(t))$. We see that for $t < 0$ but approaching $0$, that $x(t) \approx 1 + t$ by Taylor expansion, and so $f_\epsilon(x(t)) \approx f_\epsilon(1 + t)$ for $t < 0$.
On the other hand, for $t > 0$ you have $x(t) \approx 2 + 2t$, and hence for all sufficiently small $\epsilon$ we have $f_\epsilon(x(t)) = 0$ when $t = 0$.
In particular, only "half" of the bump remains.
If you carry out this computation and take the limit, indeed you will find that
$$ \lim_{\epsilon \to 0} \int f_\epsilon(x(t)) \omega(t) dt = \frac12 \omega(0) $$
and you are off by a factor of 2.
Case 2: asymmetric bumps
Now take the same $f$ as before, but define
$$ g^\pm_\epsilon(x) = \frac1\epsilon f( \frac{x-1 \pm \epsilon}{\epsilon}) $$
Note that $g^+_\epsilon$ is supported on $[1-2\epsilon,1]$ and $g^-_\epsilon$ is supported on $[1,1+2\epsilon]$.
As long as $\omega$ is a continuous function, we see that
$$\lim_{\epsilon\to 0} \int g^{\pm}(x) \omega(x) dx = \omega(1) $$
So again both of these families converge to $\delta_1$. But now, if you run the same analysis as we did before, you will find that
- $ g^+_\epsilon\circ x(t)$ converges to the distribution $\delta_0$ as $\epsilon \to 0$, while
- $ g^-_\epsilon\circ x(t)$ is always the zero distribution, for all $\epsilon$ sufficiently small.
(in fact, you can also come up with versions $h_\epsilon$ such that $h_\epsilon \circ x(t)$ converges to the distribution $\lambda \delta_0$, for any $\lambda\in (0,1)$.)
In general, if you have a piecewise smooth function $x(t)$, you can always play this sort of games with distributions $v$ with singular support at a non-smooth point of $x$.
Fundamentally, this is not too surprising: the theory of distribution is based on a linear phenomenon (pairing of functions and integrating, thereby defining a linear functional on the space of functions). Nonlinear changes of domain are not guarantee to play well with linearity. In the case where the change of variables is differentiable and regular, we know that the change is "almost linear" and so we have some hope of recovering something sensible using calculus. But as soon as you give up on this almost linearity you start running into trouble.
