A functional equation $\varphi(x) = \epsilon(x)\epsilon(x+x_1)\varphi(x+x_2)$ First, let $\epsilon:\mathbb{R}\rightarrow \mathbb{R}$ defined by $1$ over $[-1,1)$ and extended over $\mathbb{R}$ by $2$-antiperiodicity (for example, $\forall x\in [-2,-1)$, $\varepsilon(x)=-1$).
I would like to find all $2$-antiperiodic $\varphi:\mathbb{R}\rightarrow \mathbb{R}$, $x_1,x_2\in\mathbb{R}$, such that
\begin{equation}\forall x\in \mathbb{R},\quad  \varphi(x) = \epsilon(x)\epsilon(x+x_1)\varphi(x+x_2)
\end{equation}
It is assumed that $\varphi$ is piecewise-continuous, with a finite number of discontinuities over $[-1,1]$.
I managed to find some solutions as follows, but I am looking for a more general approach. By squaring the equality, it comes that $\varphi(x)^2 = \varphi(x+x_2)^2$ so $\varphi^2$ is both $2$-periodic and $x_2$-periodic. If $x_2\not\in\mathbb{Q}$, necessarily $\varphi^2$ is constant because its set of discontinuous points is not dense in $\mathbb{R}$. An example of such solution is $\varphi(x)=\epsilon(x)\epsilon(x+x_1)$, provided $x_1<2$ and $x_1+x_2$ is a multiple of $4$. Then, $\varphi(x+x_2)=\varphi(x-x_1)$, which is equal to $1$ ($\star$), hence $\varphi(x)=\varphi(x)\varphi(x+x_2)$ for $x\in[-1,1]$, and so $\varphi(x) = \epsilon(x)\epsilon(x+x_1)\varphi(x+x_2)$.
($\star$) Either $x\in[-1+x_1,1]$ and it comes directly from the definition of $\epsilon$, either $x\in[-1,-1+x_1)$ and it comes from 2-antiperiodicity of $\varphi$ and $\varepsilon$)
 A: This is only a partial answer. For $x_2$ rational, the solutions are obtained by defining $\varphi$ on some set and extending it with the constraints. We can determine which values are forced to be zero.
$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\N}{\mathbb{N}}\newcommand{\p}{\varphi}\DeclareMathOperator{\id}{id}$We work on the circle $C := \R/\Z$, the interval $[-1,3)$ becoming $[0,1)$. More generally, let $f : C \times \R \to \R$ be a function, taken here to be $f(x,y) = \epsilon(x)\epsilon(x+x_1)y$. We want to find a function $\p : C \to \R$ such that $\p(x+x_2) = f(x,\p(x))$, with a finite number of discontinuities. I replaced here $x_2$ by $-x_2$ to simplify.
Take $x \in C$ and fix $\p(x)$. The constraint gives us $\p(x+n x_2)$ for all $n \in \N$. But our choice of $\p(x)$ must be coherent with our choice of $\p(x-x_2)$, and all the choices "before". In our case, $f(x,—)$ is invertible, so it simplifies things since we ca find $\p(x-x_2)$ from $\p(x)$. (I just mention that in the general case, what we want is an element of a certain categorical filtered limit, but this limit can be empty.)
If $x_2$ is rational
If $x_2$ is rational, then let $\gcd(1,x_2)$ be the smallest non-zero multiple of $x_2$ modulo $1$. To simplify what follows, I suppose that $x_2 = \gcd(1,x_2)$, all is the same if this is not the case. We can then define $\p$ on $[0,x_2)$ arbitrarily (with a finite number of discontinuities) and extend it using $\p(x+x_2) = f(x,\p(x))$. The circle is partitioned in intervals of length $x_2$ and each iteration will give $\p$ on a new interval.
The only problem that can happen is that when we get back at the initial interval, the sign of some values may have changed. Such values must be $0$. We must use the particular value of $f$ here. You can visualize $f$ as a subset of $C$. You colorize $x \in C$ in white if $f(x,—) = \id$ and in black if $f(x,—) = -\id$. You get a symmetrical pattern, the symmetry being $x \mapsto x+1/2$. Here is a drawing of that pattern for $x_1 = 0.2$.

We partition the circle in $n$ segments of length $x_2$. To know if a point have value $0$, we take the number of equivalent points modulo $x_2$ colorized in black. If it is an odd number, the point must be $0$ (a cycle will invert the sign).
If $n$ is even
If $n$ is even, then we can write $1/2$ as a multiple of $x_2$. The point $x$ will be in the same cycle as the point $x+1/2$ and we can pair them like that inside the same cycle. Since $x$ and $x+1/2$ have the same color, there will be an even number of black points in every cycle. So we are free to choose the values we want on $[0,x_2)$ (with a finite number of discontinuities).
If $n$ is odd
If $n$ is odd, we can see what happens graphically but it is hard to describe that way here. So here is a more algebraic approach. Write $[P]$ for $1$ if $P$ is true and $0$ otherwise (they are the two elements of $\Z/2\Z$). We work modulo $1$ and when we write $x < y$, we mean it for the principal representatives.
The function $f(x,—)$ then corresponds to $[x<1/2] + [x+x_1<1/2]$. The number of inversions modulo $2$ for $x$ is
$$ g(x) := \sum_{i=0}^{n-1} [x+ix_2<1/2] + [x+ix_2+x_1<1/2] \text{.} $$
We have $\sum_{i=0}^{n-1} [x+ix_2<1/2] = [x \mod x_2 < x_2/2] + [n \equiv 3 \mod 4]$. The reason is that if $x$ is "before the cut induced by $1/2$", then the sum not modulo $2$ is $\lceil \frac{n}{2} \rceil$ and otherwise it is $\lfloor \frac{n}{2} \rfloor$.
The two copies of $[n \equiv 3 \mod 4]$ cancel out and we find
$$g(x) = [x \mod x_2 < x_2/2] + [(x+x_1) \mod x_2 < x_2/2] \text{.}$$
Graphically, it is the same pattern as $f(x,—)$.
Antiperiodicity
We now want $\p$ to be $1/2$-antiperiodic.
If $n$ is odd
Firstly, if $n$ is odd. We choose $\p$ on $[0,x_2/2)$ (respecting the constraint to be zero where $g(x) = 1$ as above). It gives the values of $\p$ on every interval $[n x_2, n x_2 + x_2/2)$. The other half-intervals are given by anti-periodicity and there is no further constraint. The reason is that $x \mapsto f(x,—)$ is $1/2$-periodic: if we start at $x$ and apply antiperiodicity to get the value at $x+1/2$, then compute the value at $x+1/2+x_1$, we get the same thing as if we first computed the value at $x+x_1$, then applied antiperiodicity to have the value at $x+x_1+1/2$. So everything is coherent.
If $n$ is even
Secondly, if $n$ is even. Forget the simplification $x_2 = \gcd(1,x_2)$. Let $k$ be such that $x_2 = k \times \gcd(1,x_2)$. We then have $\gcd(k,n) = 1$. We have $n$ intervals of length $\gcd(1,x_2)$, each colorized in black and white. The value of $\p$ on the $(i+k)$th interval (modulo $n$) depends on its value on the $i$th interval and its colorization (to know it we flip the sign or not).
How many steps $i \mapsto i+k$ must we do to make half a turn modulo $n$? It turns out to be $n/2$. Since $\gcd(k,n)=1$ and $n$ is even, $k$ must be odd and so $k \frac{n}{2} \equiv \frac{n}{2} \mod n$. But we also have $\gcd(k,n/2) = 1$, so during the displacement $i \to i+k \to i+2k \to \cdots \to i+\frac{n}{2}$, we will encounter each number once modulo $\frac{n}{2}$ (the last step $i + \frac{n}{2}$ does not count). That is to say, by symmetry of $x \mapsto f(x,—)$, we will encounter once a copy of each interval $[0,g), [g,2g), \dots, [1/2-g, 1/2)$ with $g := \gcd(1,x_2)$.
The sign of each value on the $i$th interval must be swapped when we arrive at the $(i+\frac{n}{2})$th interval. The values not swapped are forced to be zero.
Let $m := \lfloor \frac{x_1}{\gcd(1,x_2)} \rfloor$ and let $x \in [0,\gcd(1,x_2))$. Each interval fully colorized in black counts independently of $x$ and the only possibly partially colorized intreval counts if $x < x_1 \mod \gcd(1,x_2)$. So the truth-value of "$\p(x)$ must be zero" is $1+m + [x < x_1 \mod \gcd(1,x_2)]$.
If $x_2$ is irrational
We now treat the case of $x_2$ irrational. There is always the solution $\p=0$ that we rule out. As you mentioned $|\p|$ must be constant. If $x_1 = 0$, then the only solutions are the constants since we never flip the sign. Suppose $x_1 \neq 0$. The drawing for $f(x,—)$ consists of two copies of an interval semi-open. There must be a discontinuity $x$. This discontinuity will produce other discontinuities by the equation for $\varphi(x+x_2)$. The only mean to "stop" propagating that discontinuity is to hit an extremity of an interval. We must stop it in both directions, forward and backward. If we hit an open end by going forward (for instance), then we cannot hit the other by going backward because they are at a rational distance. We must hit a closed extremity backward.
Each discontinuity will be of the same type as the ones of $\epsilon$, that is $\p$ is upper semi-continuous.
We have four extremities, each of which must either annihilate a discontinuity or create one. Modulo $1$, these extremities are $0$, $x_1$, $1/2$ and $1/2+x_1$. The extremity $0$ cannot correspond to $1/2$, otherwise $x_2$ would be rational (as already said). If $0$ corresponds to $1/2+x_1$, then $x_1$ corresponds to $1/2$ and we also find that $x_2$ is rational. So $x_1$ must be an integer multiple (modulo $1$) of $x_2$.
This idea clarifies the situation a bit. For instance if $x_1 < 1/2$ and $x_2 = x_1/k$ it gives the solutions (and also in the situation of the solution you provided as an example, but I did not work it out). For more complicated values, I don't know. For particular values of $x_2$ and $x_1$, we can determine algorithmically if there is a solution by checking that the discontinuities are coherent.
A: I am thinking about your question for some time and have the following partial results. 
We’ll be consider $x_1$ and $x_2$ as given constants. 
We can split the problem putting $\varphi(x)=\delta(x)|\varphi(x)|$, where is $\delta(x)$ is the sign of $\varphi(x)$. Then the initial equation is equivalent to a system
(1) $|\varphi(x)|=|\varphi(x+x_2)|$
(2) $ \delta(x) = \varepsilon(x)\varepsilon(x+x_1)\delta(x+x_2)$. 
Assume that $x_2$ is irrational. As you have mentioned, in this case $|\varphi|=c$ is constant, so it remains to solve Equation (2). If $c=0$ then $\delta(x)=0$ for all $x$, so farther we’ll assume that $c\ne 0$. Then $\delta(x)\ne 0$ for all $x$. 
For a function $f:\Bbb R\to\Bbb R$ by $D(f)$ we denote the set of all points $x$ where $f$ is discontinuous. By the condition, $D(\varphi)$ is finite on each interval. Since $\delta(x)\ne 0$ for all $x$, $D(\delta)\subset D(\varphi)$, so the set $D(\delta)$ is finite on each interval too. Since $\delta(x)=\pm 1$ for each $x$, the problem to solve Equation (2) reduces to calculating the set $D(\delta)$.  
Let $x\in D(\delta)$. Equation (2) implies that 
(3) either $x\in D(\varepsilon)$ or $x+x_1\in D(\varepsilon)$ or $x+x_2\in D(\delta)$ 
Iterating (3), we see that either $x+x_2\in D(\delta)$  for all natural $n$, or there exists a non-negative integer $n_+(x)$ such that $x+n_+(x)x_2\in D(\varepsilon)\cup (D(\varepsilon)-x_1)$. Since $x_2$ is irrational and $\delta$ is $4$-periodic, the first case is impossible. Similarly we can show that there exists a positive integer $n_-(x)$ such that $x-n_-(x)x_2\in D(\varepsilon)\cup (D(\varepsilon)-x_1)$. Then 
$$(n_+(x)+ n_-(x))x_2\in (D(\varepsilon)\cup (D(\varepsilon)-x_1))- (D(\varepsilon)\cup (D(\varepsilon)-x_1)).$$
Recall that $D(\varepsilon)=1+2\Bbb Z$. Since $x_2$ is irrational, we have that  $x_1\not\in 2\Bbb Z$ and the only possible cases are $x+n_+(x)x_2\in D(\varepsilon)$ and $x-n_-(x)x_2\in D(\varepsilon)-x_1$ or $x-n_-(x)x_2\in D(\varepsilon)$ and $x+n_+(x)x_2\in D(\varepsilon)-x_1$.  Then 
$(n_+(x)+ n_-(x))x_2\in \pm  x_1+2\Bbb Z$, so $x_1=2k_1+k_2x_2$ for some integer numbers $k_1$ and $k_2\ne 0$. So we can specify Equation (2) as 
(4) $ \delta(x) = \varepsilon(x)\varepsilon(x+2k_1+k_2x_2)\delta(x+x_2)$.
Depending on the parity of $k_1$ it splits into 
($4_+$) $ \delta(x) = \varepsilon(x)\varepsilon(x+k_2x_2)\delta(x+x_2)$
($4_-$) $ \delta(x) =- \varepsilon(x)\varepsilon(x+k_2x_2)\delta(x+x_2)$.
I have to go to a picnic with my boss soon, so now I have no time to think farther. Maybe later I’ll manage to make some progress, but feel free to continue from here. Also Idéophage may feel free to use the above results for own answer (anyway, I am not clung to bounties. :-) )
