The first two Fourier expansions are easy: the first corresponds to the even $2L$-periodic extension of $f(x)$ while the second corresponds to the odd $2L$-periodic extension. The real challenge is the third Fourier series.

For the sake of notation, we let $\overline{f}(x)$ be the correctly extended function and $f(x) = e^x$ be an example of the original for which we wish to approximate.

The desired series
\[
\overline{f}(x) = \sum_{k=0}^\infty c_n \sin \left[ \frac{(2k+1)\pi x}{2L} \right]
\]

almost looks like the series
\[
\overline{f}(x) = \sum_{n=0}^\infty b_n \sin \left[ \frac{n \pi x}{2L} \right]
\]

but for which we have gone and set all the even modes to zero. This thus raises the question of whether we can choose a suitably extended $\overline{f}$ such that all the even modes are zero. Assuming that the integral was over $[-2L, 2L]$, recall that the computation of the modes requires
\[
b_n = \frac{1}{L} \int_0^L \overline{f}(x) \sin \left( \frac{n\pi x}{2L} \right) \, \de{x}.
\]

First, take a look at the below illustration of the base modes $\sin (n\pi x/2L)$ for the case $L = 1$.

The most important thing to notice is that when $n$ is even, the sines are odd about the line $x = L$. We want to extend our $f(x) = e^x$ so that the area of the Fourier integral on $[L, 2L$ is the negative of the area on $[0, L]$. This would work as long as $\overline{f}(x)$ is an even function about $x = L$. We thus extend our function in the following way:
\[
\overline{f}(x) = \begin{cases}
f(x) & x \in [0, L] \\
f(2L - x) & x \in [L, 2L]
\end{cases}
\]

with of course the property that $\overline{f}(x)$ is odd about $x = 0$. You can verify that this indeed creates an even mirror image of the function about $x = L$. We then have
\begin{align*}
\int_L^{2L} \overline{f}(x) \sin \left( \frac{n\pi x}{2L}\right) \de{x} &=
\int_L^{2L} f(2L - x) \sin \left( \frac{n\pi x}{2L}\right) \de{x} \\
&= -\int_L^0 f(s) \sin \left( \frac{n\pi \{2L - s\}}{2L}\right) \, \de{s} \\
&= -\int_0^L f(s) (-1)^n \sin \left( \frac{n\pi s}{2L}\right) \, \de{s}.
\end{align*}

The Fourier coefficients are then
\[
b_n = \frac{2}{2L} \int_0^{2L} \overline{f}(x) \sin \left( \frac{n\pi x}{2L}\right) \, \de{x}
= \frac{1}{L} \int_0^L f(x)(1 - (-1)^n) \sin \left( \frac{n\pi x}{2L}\right) \, \de{x},
\]

and so we are only left with the odd modes as claimed. Setting $b_{2n-1} \equiv c_n$ finishes the problem.