Solving $uu_{x_1}+u_{x_2}=u$ with boundary conditions

The characteristics written in the parameterisation invariant form is exactly as you have written

\begin{align} \frac{dx_{1}}{u} = \frac{dx_{2}}{1} = \frac{du}{u} \end{align}

Now we can solve for the characteristics. From the equality between the first and last ratios, we find

\begin{align} \frac{dx_{1}}{u} &= \frac{du}{u} \\ \implies dx_{1} &= du \\ \implies d(u - x_{1}) &= 0 \\ \implies u - x_{1} &= C_{1} \end{align}

Next is where I think you made the mistake, by simply multiplying both sides by $u$ and integrating with respect to the coordinates $x_{1}, x_{2}$ only. However, $u$ is dependent on these coordinates, meaning that

$$\int u \ dx_{2} \ne u x_{2} + C$$

From the equality between the first two ratios, we find

\begin{align} \frac{dx_{1}}{u} &= \frac{dx_{2}}{1} \\ \implies \frac{dx_{1}}{C_{1} + x_{1}} &= \frac{dx_{2}}{1} & &\text{(substitute $u = C_{1} + x_{2}$ to do the integration)} \\ \implies \ln(C_{1} + x_{1}) &= x_{2} + C_{2} \\ \implies \ln u &= x_{2} + C_{2} & &\text{(substitute back $C_{1} + x_{2} = u$ so there is only one constant in the expression)} \\ \implies u &= C_{2} e^{x_{2}} \\ \implies C_{2} &= u e^{-x_{2}} \end{align}

Combining the two relations and noting that $C_{2} = f(C_{1})$ for some arbitrary differentiable function $f$, we have that

\begin{align} u e^{-x_{2}} &= C_{2} \\ &= f(u - x_{1}) \\ \implies u &= e^{x_{2}} f(u - x_{1}) \end{align}

which you can check satisfies the PDE. Applying the initial condition $u(x_{1}, x_{2}) = 2x_{1}$ along $\Gamma := \{x_{1}, x_{2} \lvert x_{2} = 0\}$, which is the same thing as $u(x_{1},0) = 2x_{1}$, we find

\begin{align} 2x_{1} &= f(2x_{1} - x_{1}) \\ \implies f(x_{1}) &= 2x_{1} \\ \implies f(u - x_{1}) &= 2(u - x_{1}) \end{align}

so that

\begin{align} u &= 2 e^{x_{2}}(u - x_{1}) \\ \implies u &= - \frac{2 x_{1} e^{x_{2}}}{1 - 2e^{x_{2}}} \end{align}