The differential equation
$X(f) = 2f, \tag 1$
where $X$ is the vector field
$X = x_1^2 \dfrac{\partial}{\partial x_1} -
x_2^2 \dfrac{\partial}{\partial _2}, \tag 2$
may also be written in the form
$\dfrac{\partial f}{\partial t} = 2f, \tag 3$
where $t$ is the running parameter along the integral curves of $X$; here we make the usual identification
$X \equiv \dfrac{\partial}{\partial t}. \tag 4$
It will be noted that in fact (3) is simply an ordinary differential equation; thus, along any trajectory of $X$, we may in the usual manner write
$X(\ln f) = \dfrac{\partial (\ln f)}{\partial t} = \dfrac{1}{f} \dfrac{\partial f}{\partial t} = 2, \tag 5$
which may be integrated 'twixt $t_0$ and $t$ to yield
$\ln \left ( \dfrac{f(t)}{f(t_0)} \right ) = \ln(f(t) ) - \ln(f(0)) = 2(t - t_0), \tag 6$
or
$f(t) = f(t_0)e^{2(t - t_0)}, \tag 7$
which expresses the evolution of $f$ along any trajectory of $X$ in terms of the parameter $t$ such that (4) binds. The reader will no doubt recognize that (7) presents
$f:\Bbb R \to \Bbb R \tag 8$
as a function of the single parameter $t$, whereas the question specifies that
$f:\Bbb R^2 \to \Bbb R \tag 9$
is indeed dependent upon the two variables $x_1, x_2$; in reconciling these dual points of view we will exploit the fact that, along the integral curves of $X$ the coordinates $x_1, x_2$ must satisfy the differential equations
$\dot x_1 = x_1^2, \tag{10}$
$\dot x_2 = -x_2^2; \tag{11}$
these equations are both of the general form
$y = ay^2, \tag{12}$
and the solution is derived below in an appendix to this answer. We in fact have:
$x_1(t) = x_{10}(1 - x_{10}(t - t_0))^{-1}, \; x_1(t_0) = x_{10}, \tag{13}$
$x_2(t) = x_{20}(1 + x_{20}(t - t_0))^{-1}, \; x_2(t_0) = x_{20}; \tag{14}$
then what we have written as
$f(t) = f(x_1(t), x_2(t)), \tag{15}$
and we may find $f(x_1, x_2)$ for arbitrary $x_1, x_2$ by discovering an $x_{10}, x_{20}$, $t$ and $t_0$ (if indeed such concurrently exist) such that (13) and (14) bind, where $x_{10}$ and $x_{20}$ are the coordinates of a point at which $f(t_0)$ is specified; typically, $x_{10}$, $x_{20}$ will lie in some submanifold, in the present instance in fact a curve in $\Bbb R^2$ that is, apparently, the circle $(\cos \theta, \sin \theta)$ on which we have
$f(\theta) = \cos \theta + \sin \theta; \tag{16}$
we may, via (13) and (14), express $x_1$, $x_2$ by means of a coordinate transformation which gives them in terms of $t$ and $\theta$:
$x_1(t, \theta) = \cos \theta (1 - \cos \theta (t - t_0))^{-1}, \tag{17}$
$x_2(t, \theta) = \sin \theta (1 + \sin \theta (t - t_0))^{-1}; \tag{18}$
in these coordinates we have, by (7),
$f(t, \theta) = e^{2(t - t_0)} (\cos \theta + \sin \theta); \tag{19}$
in principle, the transformation (17)-(18) may be reversed; in so doubg, the identity
$\sin^2 \theta + \cos^2 \theta = 1 \tag{20}$
may prove useful, allowing as it does the expression $\sin \theta$ in terms of $\cos \theta$.
Appendix:
$\dot y = ay^2, \; y(t_0) = y_0; \tag 1$
$y^{-2}\dot y = a; \tag2$
$y_0^{-1} - y^{-1} = \displaystyle \int_{y_0}^y y^{-2}dy = \int_{t_0}^t a \; ds = a(t - t_0); \tag 3$
$y^{-1} = y_0^{-1} - a(t - t_0) = y_0^{-1} - y_0^{-1} y_0 a(t - t_0) = y_0^{-1}(1 - y_0 a(t - t_0));\tag 4$
$y = y_0(1 - y_0 a(t - t_0))^{-1} = \dfrac {y_0}{1 - y_0 a (t - t_0)}; \tag 5$
we apply these calculations to find (locally) the integral curves of the vector field
$X = x_1^2 \dfrac{\partial}{\partial x_1} -
x_2^2 \dfrac{\partial}{\partial _2} \tag 6$
with initial condition
$x_1(t_0) = x_{10}, \; x_2(t_0) = x_{20}; \tag 7$
the formula (5) applies to this situation when we take
$a = 1, \; y = x_1; \; a = -1, y = x_2; \tag 8$
we have
$x_1(t) = x_{10}(1 - x_{10}(t - t_0))^{-1}, \tag 9$
$x_2(t) = x_{20}(1 + x_{20}(t - t_0))^{-1}, \tag{10}$