Showing that such a homomorphism exists I'm looking to show that there is a homomorphism $f : S_n\to C_2$, where $C_2 = \langle a\ |\ a^2=1\rangle$, and $f$ takes transpositions to $a$ (a transposition being a 2-cycle $(i_1, i_2)$).
The way I attempted to do it is to simply "choose" $f$ to be a multiplicative function from $S_n$ to $C_2$ taking transpositions to $a$, and showing that such a function is well defined. Since any permutation can be written as the product of disjoint cycles, and each cycle can be written as the product of transpositions, we can define $f(\sigma)$ for each $\sigma$, and then... I don't really know?
I get very confused about arguments like these. How do we show that such a homomorphism exists? Where do we start, what assumptions do we start with, and what are we "allowed" to say? Are we allowed to choose $f$ to be multiplicative, or are we just supposed to use the assumption that $f$ takes transpositions to $a$? Are we allowed to assume that $f$ is explicitly a homomorphism that takes transpositions to $a$, and then simply show that such a map is well defined? What is the process of showing that "a homomorphism with these properties exists" in general? I'm just generally very confused...
 A: Basically you can assume whatever you can prove (restricted to whatever your instructor allows you to do).  For example, if you can prove that you can write a permutation as a matrix (which involves checking that any permutation only has one matrix representation and that they compose the way you expect), then you can certainly use it.
To follow the approach given in the question a bit more, saying "define $f$ to be a multiplicative function that sends transpositions to $a$" is the same as saying "there exists a unique function $f$ that is multiplicative and sends transpositions to $a$".  For this problem, there is no reason to check uniqueness, because the existence of such an $f$ alone gives you a homomorphism.  
The step about splitting any permutation $\sigma$ into disjoint cycles gives you existence.  The way it does this is by defining $f$ for each permutation uniquely; once you check that this is multiplicative and sends transpositions to $a$, you have the existence of such an $f$.
It is important to note that before one can use cycle decomposition, you must prove that each permutation has a unique cycle decomposition; furthermore, I expect that when proving that $f$ is multiplicative, you will want to prove some rule about how cycle decompositions compose, just as when you were trying to use the matrix representation of a permutation.
A: The following is a pretty lengthy and naive way of answering this, but this approach may be helpful for building an understanding of what's going on. You can always define a function however you want, as long as it's a properly defined function. Then you need to prove any properties that it may have (that it's a homomorphism, for example), but you can always do rough work with it under some assumptions as long as you prove the assumptions afterwards.
So, typically, you construct a function and then prove its well-defined and a homomorphism, but you can always do rough work beforehand.
Rough work
We do some rough work first. Using your train of thought, consider some permutation $\sigma \in S_{n}$ that is not a transposition. Then it is either even or odd (a fact that you'd have to prove or be allowed to use without proof), so we'll consider each separately:
If $\sigma$ is even, then it is the product of an even number of transpositions, so say there are transpositions $\tau_{i}$ such that
$$
\sigma = \tau_{1}\tau_{2}\cdots\tau_{2k-1}\tau_{2k}
$$
for $k \in \mathbb{N}$. Certainly these exist and are in $S_{n}$. What's the image under $f$ of such a product? Well, observe that if $f$ were a homomorphism we would have
$$
f(\sigma) = f(\tau_{1}\tau_{2}\cdots\tau_{2k-1}\tau_{2k}) = f(\tau_{1})f(\tau_{2})\cdots f(\tau_{2k-1})f(\tau_{2k}).
$$
But each transposition maps to $a$, so
$$
f(\sigma) = a\cdot a\cdots a \cdot a = a^{2k} = (a^{2})^{k} = 1^{k} = 1.
$$
Thus, if $\sigma$ is an even permutation, it maps to the identity.
Similarly, if $\sigma$ is an odd permutation, write
$$
\sigma = \tau_{1}\tau_{2}\cdots\tau_{2k}\tau_{2k+1}
$$
for $k \in \mathbb{N}$. The image of this is
$$
f(\sigma) = f(\tau_{1}\tau_{2}\cdots\tau_{2k}\tau_{2k+1}) = f(\tau_{1})f(\tau_{2})\cdots f(\tau_{2k})f(\tau_{2k+1})
$$
and since each transposition maps to $a$, we have
$$
f(\sigma) = a\cdot a\cdots a \cdot a = a^{2k+1} = (a^{2})^{k}\cdot a = 1 \cdot a = a.
$$
Thus, if $\sigma$ is an odd permutation, it maps to $a$.
Working under the assumption that $f$ were a homomorphism, we see that $f$ maps even permutations to $1$, and odd permutations to $a$. Let's now formally define $f$ and prove that $f$ is a homomorphism.
Proof
Define $f : S_{n} \to C_{2}$ by
$$
f(\sigma) = \begin{cases} 1, & \text{if $\sigma$ is even,} \\ a, & \text{if $\sigma$ is odd.} \\ \end{cases}
$$
We need to show that $f$ is a homomorphism. Let $\sigma_{1}, \sigma_{2} \in S_{n}$ be even permutations, and let $\gamma_{1},\gamma_{2} \in S_{n}$ be odd permutations. Then $\sigma_{1}\sigma_{2}$ and $\gamma_{1}\gamma_{2}$ are even (check), so
$$
f(\sigma_{1}\sigma_{2}) = 1 = 1 \cdot 1 = f(\sigma_{1})\cdot f(\sigma_{2}) \quad\text{and}\quad f(\gamma_{1}\gamma_{2}) = 1 = a \cdot a = f(\gamma_{1})\cdot f(\gamma_{2}).
$$
Similarly, if $\sigma \in S_{n}$ is even and $\gamma \in S_{n}$ is odd, then $\sigma\gamma$ and $\gamma\sigma$ are both odd (check), so
$$
f(\sigma\gamma) = a = 1 \cdot a = f(\sigma)\cdot f(\gamma) \quad\text{and}\quad f(\gamma\sigma) = a = a \cdot 1 = f(\gamma)\cdot f(\sigma).
$$
Thus, for any product of permutations $\sigma\gamma$ for $\sigma, \gamma \in S_{n}$, we have that
$$
f(\sigma\gamma) = f(\sigma) \cdot f(\gamma)
$$
and thus $f$ is a homomorphism. Clearly $f$ takes transpositions to $a$, as every transposition is an odd permutation.
