So the wave equation:
$$U_{tt}-c^2U_{xx}=0.$$
This is as far as I've gotten so far:
Let $a:=x+k, b:=t+h$, then by chain rule
\begin{equation} \frac{\partial U}{\partial x}=\frac{\partial U}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial U}{\partial b}\frac{\partial b}{\partial x}, \end{equation} \begin{equation} \frac{\partial U}{\partial t}=\frac{\partial U}{\partial a}\frac{\partial a}{\partial t}+\frac{\partial U}{\partial b}\frac{\partial b}{\partial t}. \end{equation} Using the alternate notation for partial derivatives, we have \begin{equation} U_x=U_a(1)+U_b(0)=U_a, \end{equation} \begin{equation} U_t=U_a(0)+U_b(1)=U_b. \end{equation} Now, since we require a second derivative, we must differentiate (31) and (32) once again. We have \begin{equation} \frac{\partial^2 U}{\partial x^2}=\frac{\partial}{\partial x}\Big(\frac{\partial U}{\partial x}\Big)=\frac{\partial}{\partial x}\Big(\frac{\partial U}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial U}{\partial b}\frac{\partial b}{\partial x}\Big), \end{equation} or equivalently, by the product rule \begin{equation} \frac{\partial^2 U}{\partial x^2}=\frac{\partial U}{\partial a}\frac{\partial}{\partial x}\Big(\frac{\partial a}{\partial x}\Big)+\frac{\partial a}{\partial x}\frac{\partial}{\partial x}\Big(\frac{\partial U}{\partial a}\Big)+ \frac{\partial U}{\partial b}\frac{\partial}{\partial x}\Big(\frac{\partial b}{\partial x}\Big)+\frac{\partial b}{\partial x}\frac{\partial}{\partial x}\Big(\frac{\partial U}{\partial b}\Big), \end{equation}
however it then just unravels into a big mess, and would like a pointer or two on how to proceed from here.
OK, so we have
$$\frac{\partial a}{\partial x}=1, \frac{\partial b}{\partial x}=0, \frac{\partial a}{\partial t}=0, \frac{\partial b}{\partial t}=1,$$
then, our second derivative equation simplifies to
\begin{equation} U_{xx}=\frac{\partial}{\partial x}\Big(\frac{\partial U}{\partial a}\Big), \end{equation}
using the fact that $U_x=U_a$ as done above, we can say $$U_{xx}=\frac{\partial}{\partial a}\frac{\partial U}{\partial a}=\frac{\partial^2}{\partial a^2}=U_{aa}.$$
Similarly, for $U_{tt}$ \begin{equation} \frac{\partial^2 U}{\partial t^2}=\frac{\partial}{\partial t}\Big(\frac{\partial U}{\partial t}\Big)=\frac{\partial}{\partial t}\Big(\frac{\partial U}{\partial a}\frac{\partial a}{\partial t}+\frac{\partial U}{\partial b}\frac{\partial b}{\partial t}\Big), \end{equation} or equivalently, by the product rule \begin{equation} \frac{\partial^2 U}{\partial t^2}=\frac{\partial U}{\partial a}\frac{\partial}{\partial t}\Big(\frac{\partial a}{\partial t}\Big)+\frac{\partial a}{\partial t}\frac{\partial}{\partial t}\Big(\frac{\partial U}{\partial a}\Big)+ \frac{\partial U}{\partial b}\frac{\partial}{\partial t}\Big(\frac{\partial b}{\partial t}\Big)+\frac{\partial b}{\partial t}\frac{\partial}{\partial t}\Big(\frac{\partial U}{\partial b}\Big), \end{equation} which simplifies to \begin{equation} \frac{\partial^2 U}{\partial t^2}=\frac{\partial}{\partial t}\frac{\partial U}{\partial b}, \end{equation} now, using the relation established with our first order derivatives, we get \begin{equation} U_{tt}=\frac{\partial^2 U}{\partial t^2}=\frac{\partial}{\partial b}\frac{\partial U}{\partial b}=\frac{\partial^2 U}{\partial b^2}=U_{bb}. \end{equation} Then \begin{equation} U_{tt}-c^2U_{xx}=0 \Rightarrow U_{bb}-c^2U_{aa}=0, \end{equation} as desired.
This looks good.
I'd propose a couple of minor simplifications/clarifications: First, the differential equation is really, $$ (\mbox{second derivative in time}) - c^2 (\mbox{second derivative in space}) = 0 $$ So by the definition of $a$ and $b$, we have $U_{bb} - c^2 U_{aa} = 0$.
Second, to avoid confusion, consider renaming $U(x+k, t+h)$ to be something like $F$. This will simplify your notation -- see below. Your goal is now to show that $F_{tt} - c^2 F_{xx} = 0$.
Third, I'd suggest simplifying your derivatives each step of the differentiation. That will make it easier to track. For example: $$ F_x = U_aa_x + U_bb_x = U_a $$ and so $$ F_{xx} = (F_x)_x = (U_a)_x = U_{aa}a_x + U_{ab}b_x = U_{aa} $$ and similarly with differentiation in time: $$ F_t = U_aa_t + U_bb_t = U_b,\ \ F_{tt} = (F_t)_t = (U_b)_t = U_{ba}a_t + U_{bb}b_t = U_{bb} $$
The denouement is simply $$ F_{xx} - c^2 F_{tt} = U_{aa} - c^2 U_{bb} = 0. $$