Green's Function / Impulse Response Confusion

You are simply reading it backwards. When they say "impulse response of an inhomogeneous linear differential equation defined on a domain, with specified initial conditions or boundary conditions" they mean the function $G(x,s)$ such that $L(G(x,s)) = \delta(x-s)$.

I know it is a little counterintuitive at first since one thinks on the differential operator $L$ as something that you feed with a function and gives back a function. But in this case the operator we are dealing with is in fact the solution operator of the differential equation. That is $S:X\rightarrow Y$, where $X$, $Y$ are function spaces such that $S(f) = u$ when \begin{equation} \begin{cases} Lu=f &\text{ in } \Omega\\ u = 0 &\text{ in } \partial\Omega \end{cases} \text{ (It can also be a different boundary condition depending on the problem)} \end{equation}

Think of the Green functions and the $\delta$ in the following way to notice why this is useful, the $\delta$ is "kind of a base of the functions spaces" since you can "write" any function as \begin{gather} f(x)"=" \sum_s f(s)\delta(x-s)\\ \text{ (It really is an integral not a sum, in fact is a convolution integral)} \end{gather} And, since the solution operator is linear, then to solve the problem for a general $f$ it is enough to solve it for the deltas and then you just sum using superposition. The Green functions are just the solutions of the deltas, that is \begin{equation} G(x,s) = S(\delta(x-s))\\ \end{equation} so \begin{equation} u(x) = S(f)(x) "=" \sum_s f(s)S(\delta(x-s)) = \sum_s f(s)G(x,s) \end{equation}

Notice that $s$ is a parameter not the variable of the delta function so $f(s)$ is a constant for the solution operator. Notice also that this is not a really formal answer, but I hope it is useful to star understanding what the Green functions are about.