Consider the function $f(a,b) = u(t-a)-u(t-b); a>b$. This function is 1 in $[a,b)$ and 0 elsewhere.
So, suppose you want to write
$$g(x) =
\begin{cases}
g_1(x), & \text{if } a_1<x<a_2 \\
g_2(x), & \text{if } a_2<x<a_3\\
\ldots
\end{cases}$$
Then,
$$g(x) = g_1(x)f(a_2,a_1) + g_2(x)f(a_3,a_2)+\ldots$$
You can solve for $u$ in a simpler fashion as follows. Since your PDE is linear, it must hold:
$$\frac{dt}{1} = \frac{dx}{c} = \frac{du}{e^{2x}},$$
which gives you the characteristics:
$$ \begin{align}
dx - cdt = & 0 \iff x- ct = \eta, \\
e^{2x}dx/c = & du \iff u = e^{2x}/(2c) + \xi.
\end{align}$$ Since the original equation is a 1st order PDE, then $\xi$ must be a function of $\eta$, $\xi = g(\eta)$ (or conversely, $\eta = h(\xi)$), so your solution is given by (prove that it satisfies the original PDE):
$$u(t,x) = e^{2x}/(2c) + g(x-ct).$$
Apply now the initial condition to obtain:
$$u(0,x) = f(x) = e^{2x}/(2c) + g(x),$$
so $g(x) = f(x)-e^{2x}/(2c)$, and hence the solution which satisfies the given initial condition is:
$$u(t,x) = e^{2x}/(2c) + f(x-ct) - e^{2(x-ct)}/(2c).$$
Cheers!
Edit
Everything commented above is valid for $c \neq 0$ (non-zero phase velocity). What happens if $c = 0$? Then, the original equation can be rewritten as follows:
$$u_t = e^{2x},$$
so the solution can be easily achieved by integrating "partially" with respect to $t$:
$$u(x,t) = e^{2x} t + h(x),$$
being $h(x)$ an arbitrary function. Apply again the initial condition to obtain $f(x) = h(x)$, so the solution becomes:
$$u(x,t) = e^{2x} t + h(x).$$
(Note that you could have done: $\int^{u(t,x)}_f u_t \, dt = \int^t_0 e^{2x} \, dt$ or write again the characteristics equations, which yields $dx = 0 \Rightarrow x = \eta$).
Best Answer
The sum of functions is defined pointwise. For example, say you have two step functions
$u_1(t) = \begin{cases} 0, & t < 1 \\ 1, & t \ge 1 \end{cases}$
and
$u_2(t) = \begin{cases} 0, & t < 2 \\ 1, & t \ge 2 \end{cases}$.
Then a linear combination of them, say $au_1(t) + bu_2(t)$ would be
$au_1(t) + bu_2(t) = \begin{cases} 0, & t < 1 \\ a, & 1 \le t < 2 \\ a+b, & t \ge 2 \end{cases}$.