It sounds like the functions you're dealing with are pretty nice and admit Laplace transforms. So we know that for two functions $f$ and $g$, the Laplace transform of their "convolution" (as written below) is:

$\mathcal{L}(f\star g)=\mathcal{L}\int_0^t f(t-x)g(x)dx=F(s)G(s)$

where $F$ and $G$ are the Laplace transforms of $f,g$ respectively. Look at the proof in this link here. For $f(2t-x)$, following the notation in the link and working bottom up, the only difference that occurs is that instead of $t=\sigma+\tau$, you replace with $2t=\sigma+\tau$. This gives

$\mathcal{L}\int_0^t f(2t-x)g(x)dx=\frac{1}{2}\int_0^\infty \int_0^\infty f(\sigma)e^{-s(\sigma+\tau)/2}d\sigma g(\tau)d\tau=\frac{1}{2}F(s/2) G(s/2)$

So for your equation, you get:

$H(s)=\frac{1}{2}F(s/2)G(s/2)$

and you can now solve for $g$ by using the Laplace transform inverse.

It sounds like the functions you're dealing with are pretty nice and admit Laplace transforms. So we know that for two functions $f$ and $g$, the Laplace transform of their "convolution" (as written below) is:

$\mathcal{L}(f\star g)=\mathcal{L}\int_0^t f(t-x)g(x)dx=F(s)G(s)$

where $F$ and $G$ are the Laplace transforms of $f,g$ respectively. Look at the proof in this link here. For $f(2t-x)$, following the notation in the link and working bottom up, the only difference that occurs is that instead of $t=\sigma+\tau$, you replace with $2t=\sigma+\tau$. This gives

$\mathcal{L}\int_0^t f(2t-x)g(x)dx=\frac{1}{2}\int_0^\infty \int_0^\infty f(\sigma)e^{-s(\sigma+\tau)/2}d\sigma g(\tau)d\tau=\frac{1}{2}F(s/2) G(s/2)$

So for your equation, you get:

$H(s)=\frac{1}{2}F(s/2)G(s/2)$

and you can now solve for $g$ by using the Laplace transform inverse.

As far as references go, try "Introduction to integral equations with applications" By Abdul J. Jerri.