Thanks to the formula $e^{ix} = \cos(x) + i \sin(x)$ you can transform
$$\sum_{n = -\infty}^\infty c_n(l) e^{in\pi x \, /\, l}$$
into the following:
$$\sum_{n = -\infty}^\infty c_n(l) \cos( n \pi x \, / \, l) + \sum_{n = -\infty}^\infty i c_n(l) \sin(n\pi x\,/\,l).$$
Then, by exploiting the identities $\cos(-x)=\cos(x)$ and $\sin(-x) = -\sin(x)$, you get
$$c_0(l) + \sum_{n = 1}^\infty [c_n(l)+c_{-n}(l)] \cos( n \pi x \, / \, l) + \sum_{n = 1}^\infty i [c_n(l)-c_{-n}(l)] \sin(n\pi x\,/\,l) $$
If the coefficients satisfy a simple relationship, the series will have a real value: since
$$ \cos{x} = \frac{e^{ix}+e^{-ix}}{2}, \qquad \sin{x} = \frac{e^{ix}-e^{-ix}}{2i} $$
(this comes from two instances of Euler's formula, for example), we have
$$ a_n \cos{nx}+b_n\sin{nx} = a_n \frac{e^{inx}+e^{-inx}}{2} + b_n \frac{e^{inx}-e^{-inx}}{2i} = \frac{a_n-ib_n}{2} e^{inx} + \frac{a_n+ib_n}{2} e^{-inx}. $$
In particular, one coefficient is the complex conjugate of the other. Inverting this, if we have a Fourier series
$$ \sum_{n=-\infty}^{\infty} c_n e^{inx}, $$
the series is real if and only if $c_{-n} = c_n^*$ (and we find $a_n = c_n+c_{-n} $, $ b_n = i(c_n-c_{-n}) $ ).
Why would we write it as complex exponentials? Firstly, $\int_{-\pi}^{\pi} e^{inx} f(x) \, dx$ is often easier to calculate than the trigonometric forms, secondly, we only have to do one integral instead of (often) three: one for the constant term, one for the cosine part and one for the sine part, and thirdly, many other important formulae are simpler written in terms of the $c_n$: for example, Parseval/Plancherel is
$$ \frac{1}{2\pi}\int_{-\pi}^{\pi} \lvert f(x) \rvert^2 \, dx = \sum_{n=-\infty}^{\infty} \lvert c_n \rvert^2 $$
instead of
$$ \frac{1}{2\pi}\int_{-\pi}^{\pi} \lvert f(x) \rvert^2 \, dx = \frac{a_0^2}{4} +\frac{1}{2}\sum_{n=1}^{\infty} a_n^2 + b_n^2 $$
That's not to say the real form doesn't have its uses: if we want to keep everything real, we use it (particularly in applications like solving Laplace's equation, for example), or if the function is clearly odd or even, we only have to calculate one set of coefficients, and the work is effectively halved.
Best Answer
$\newcommand{\Vector}[1]{\mathbf{#1}}\newcommand{\vece}{\Vector{e}}$The linked questions provide good answers, but may be at the technical end of "intuitive". Here's a fast-and-loose conceptual motivation:
If $\bigl(V, \langle\ ,\ \rangle\bigr)$ is an $N$-dimensional real inner product space, and if $\{\vece_{n}\}_{n=1}^{N}$ is an (ordered) orthonormal basis, then an arbitrary vector $v$ in $V$ may be written as a linear combination $$ v = \sum_{n=1}^{N} \langle v, \vece_{n}\rangle \vece_{n}. \tag{1} $$ Indeed, $\{\vece_{n}\}$ is a basis of $V$, so there exist real coefficients $a_{k}$ such that $$ v = \sum_{k=1}^{N} a_{k} \vece_{k}. \tag{2} $$ Taking the inner product of each side with $\vece_{n}$ gives $\langle v, \vece_{n}\rangle = a_{n}$ because the basis $\{\vece_{n}\}$ is orthonormal.
Loosely, one might expect a similar conclusion to hold if $V$ is infinite-dimensional. Getting the definitions and hypotheses right, and proving a version of (1) in this new setting, is why any "honest" answer is bound to be technical. Phrases in quotes below are not mathematically correct, and therefore require careful inspection and/or justification.
Intuitively, let $L > 0$ be real, let $V$ be "the space of real-valued functions" on $[-L, L]$, and define an "inner product" by $$ \langle f, g\rangle = \frac{1}{L} \int_{-L}^{L} f(t) g(t)\, dt. $$ The functions $$ C_{n}(t) = \begin{cases} 1/\sqrt{2}, & n = 0, \\ \cos(n\pi t/L), & n > 0; \end{cases}\qquad S_{n}(t) = \sin(n\pi t/L),\quad n > 0; $$ turn out (by elementary calculus and trigonometry) to form an "orthonormal basis" of $V$.
Loosely, we expect that if $f$ is a function, we can express $f$ as an infinite sum of these basis functions, and the coefficients are the inner products of $f$ with the basis elements, i.e. (for $n > 0$), \begin{align*} a_{0} &= \langle f, 1\rangle = \frac{1}{L} \int_{-L}^{L} f(t)\, dt, \\ a_{n} &= \langle f, C_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \cos(n\pi t/L)\, dt, \\ b_{n} &= \langle f, S_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \sin(n\pi t/L)\, dt, \\ f(t) &= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n} C_{n}(t) + b_{n} S_{n}(t), \\ &= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos(n\pi t/L) + b_{n} \sin(n\pi t/L). \end{align*} (The "special" factor of $1/2$ on the constant term arises because $C_{0} = 1/\sqrt{2} \neq 1$.)