I want to show that $$w(x,t)=\frac{1}{2c}\int_0^{t-\frac{x}{c}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)\,dy\,d\tau+\frac{1}{2c}\int_{t-\frac{x}{c}}^t\int_{x-c(t-\tau)}^{x+c(t-\tau)}f(y,\tau)\,dy\,d\tau$$ is the solution of the problem $$w_{tt}=c^2w_{xx}+f(x,t) , \ \ x>0, t>0 \\ w(x,0)=w_t(x,0)=0, \ \ x>0 \\ w(0,t)=0 , \ \ t\geq 0$$ I am checking if the condition $w(x,0)=0$ holds and I get the following: \begin{align*}w(x,0)&=\frac{1}{2c}\int_0^{0-\frac{x}{c}}\int_{c(0-\tau)-x}^{x+c(0-\tau)}f(y,\tau)\,dy\,d\tau+\frac{1}{2c}\int_{0-\frac{x}{c}}^0\int_{x-c(0-\tau)}^{x+c(0-\tau)}f(y,\tau)\,dy\,d\tau\\ & =\frac{1}{2c}\int_0^{-\frac{x}{c}}\int_{-c\tau-x}^{x-c\tau}f(y,\tau)\,dy\,d\tau+\frac{1}{2c}\int_{-\frac{x}{c}}^0\int_{x+c\tau}^{x-c\tau}f(y,\tau)\,dy\,d\tau\\ & = -\frac{1}{2c}\int_{-\frac{x}{c}}^0\int_{-c\tau-x}^{x-c\tau}f(y,\tau)\,dy\,d\tau+\frac{1}{2c}\int_{-\frac{x}{c}}^0\int_{x+c\tau}^{x-c\tau}f(y,\tau)\,dy\,d\tau\\ & = \frac{1}{2c}\int_{-\frac{x}{c}}^0\left (-\int_{-c\tau-x}^{x-c\tau}f(y,\tau)\,dy\,d\tau+\int_{x+c\tau}^{x-c\tau}f(y,\tau)\,dy\,d\tau\right )\\ & = \frac{1}{2c}\int_{-\frac{x}{c}}^0\left (\int_{x-c\tau}^{-c\tau-x}f(y,\tau)\,dy\,d\tau+\int_{x+c\tau}^{x-c\tau}f(y,\tau)\,dy\,d\tau\right )\\ & = \frac{1}{2c}\int_{-\frac{x}{c}}^0\left (\int_{x+c\tau}^{-c\tau-x}f(y,\tau)\,dy\,d\tau\right )\end{align*}
How do we get that this is equal to $0$ ?
The first formula you wrote does not make sense as it is, because $f=f(x, t)$ is defined only for $x\in (0, \infty)$. However you are implicitly using a standard trick, that of considering $f(x, t) = -f(-x, t)$ if $x\le 0$. (This is called reflection trick or something like that). In particular, the integral $$ \int_{x+c\tau}^{-x-c\tau} f(y, s)\, dy= 0, $$ because it is the integral of an odd function over an interval that is symmetric around $0$.