I (think I) understand everything up until the step that integrates $(f(x))^2$ at $x \in [-\pi, \pi]$.
I understand why the zero occurred, my understanding is that $\cos$ and $\sin$ cancel out after the integration since the summation of $m$ and $n$ mounts the function to zero. An indefinite integration check solidifies this suspicion.
The main problem I have is that after the integral has resolved, is where does the delta function come from. There is no literature mentioning of this. From my understanding of integral, this means that
$$f(x) = \cos(nx)\cdot\cos(mx)$$
and
$$f'(x) = \pi\cdot\delta_{nm}.$$
The $\delta$ being the Dirac delta function. I feel like asking this problem will give me a more proper understanding of the Dirac delta function.
The delta comes from a quirk about the $\sin(nx)$ and $\cos(nx)$ functions: they're all mutually orthogonal. This means that, when taking the inner product $$\int_{-\pi}^\pi \sin(nx)\sin(mx) \mathrm{d}x,$$ the result is zero if $n \neq m$, or something strictly positive (in this case, $\pi$) otherwise. This is similarly true for $\cos$. Also, taking a mixed integral $$\int_{-\pi}^\pi \sin(nx)\cos(mx) \mathrm{d}x$$ always results in $0$, as the product is odd.
You can verify these facts by using the following formulae: \begin{align*} \cos(\theta)\cos(\phi) &= \frac{1}{2}(\cos(\theta - \phi) + \cos(\theta + \phi)), \\ \sin(\theta)\sin(\phi) &= \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi)). \end{align*} If we take $\theta = mx$ and $\phi = nx$, then $$\cos(mx)\cos(nx) = \frac{1}{2}(\cos((m - n)x) + \cos((m + n)x).$$ Note that $\cos((m + n)x)$ is a (possibly high-frequency) sine wave that spends as much time over the $x$-axis as under it, so its integral will be $0$. The same is true for $\cos((m - n)x)$, unless $m = n$, in which case you are integrating the constant function $1$ over $-\pi$ to $\pi$. Hence, the integral will be $\pi$ if $m = n$ or $0$ otherwise.