For a map $P^1 \to P^1$, the possible ramifications are limited by Riemann-Hurwitz and ... For a map $\phi : P^1 \to P^1$, the possible ramifications are limited by Riemann-Hurwitz (here saying that $2(deg \phi - 1) = \Sigma_{P \in P^1} (e_\phi(P) - 1)$) and the formula $\Sigma_{P \in \phi^{-1}(Q)} e_\phi(P) = \deg \phi$.
For example, if the degree of $\phi$ is 2, the only possibility is that there are two points of ramification, each ramified to order 2. Let's represent this by the ordered pair $(2,2)$. One can easily write down such a map, or guarantee its existence from Riemann Roch (a function in the space of global sections of the degree zero divisor $2P - 2Q$).
For higher degree maps, we can write down similar (finite) lists of possibilities: how many points are ramified, and what their ramifications are.
If $\deg = 3$, we have $(3,3)$ (which means two points in the domain are triply ramified), and $(2, 2, 2, 2)$ (meaning there are four points in the domain of ramification 2, each one is paired with some other unramified point that has the same image, everywhere else it is a 3-1 cover). The first one is easy to write down, the latter one I am unsure about. 
My question is: Is there a function realizing $(2,2,2,2)$? Is there a function realizing every tuple arising in this way? Maybe it follows from Riemann-Roch somehow?
More thoughts about this:
It also seems like one could approach this via the function field, by adjoining n-th roots of elements of the form $t - a$ to try to produce prescribed ramification there. It's not at all clear to me what the ramification of the resulting map would look like - it should be interesting for me to work this out. 
 A: Every ramification profile that satisfies the two conditions you stated (Riemann-Hurwitz and the degree bound on the ramification index) is possible.
Sketch of proof: the set of maps $\phi : \mathbb{P}^1 \to \mathbb{P}^1$ of degree $d$ is a Grassmannian $Gr(2,d+1)$: writing $\phi = \frac{f(z)}{g(z)}$ with $f(z),g(z)$ polynomials of degree $d$, we get a rank-2 subspace
$$\mathrm{span}(f(z),g(z)) \subset \{\text{degree $\leq d$ polynomials in } z\}.$$
In fact this space is unaffected by changing coordinates on the target $\mathbb{P}^1$. Conversely, given a rank-2 subspace, we can choose any basis for it to use as $f$ and $g$. This reconstructs $\phi$ up to changing coordinates on the target, which doesn't affect the ramification profile.
The condition "$\phi$ is (at least) simply ramified at $p$" is a codimension one condition on this $Gr(2,d+1)$, a divisor. This divisor is ample (it is a hyperplane section of $Gr(2,d+1)$ in the Plücker embedding), which means every positive-dimensional subvariety of $Gr(2,n)$ will intersect it.
This is enough to know that the profile of "all simply-ramified points" must be achievable: just intersect each of the subsets
$$\{\phi \text{ ramified at } p \} \subset Gr(2,d+1)$$
for the distinct points $p$. (Note that the dimension of $Gr(2,d+1)$ is $2(d-1)$, so we can force exactly the right number of points to be simply ramified.)
To get specific other profiles, the story is a morally similar but technically a little different -- I'm not sure if there's an easier answer than to use some amount of Schubert calculus / combinatorics. This is a computation in the cohomology ring $H^*(Gr(2,d+1))$ of $Gr(2,d+1)$, namely multiplying the classes of several of what are called (special) `Schubert varieties' -- these are very well studied and understood.
In this case, the classes corresponding to "having a specified ramification index at $p$" will always multiply to something nonzero in $H^*(Gr(2,d+1))$, which in particular implies that the locus corresponding to $\phi$ having a specified ramification profile is always nonempty.
For more sources, search for "the Wronski problem".
A: This is wrong, as Jake Levinson points out.
So there is a simple minded answer to my question, which is no:
This is just because a rational (hence regular) function on $P^1$ is determined by its poles and zeros, so not all combinations of ramifications are possible.
Thanks Jake, I found my mistake:
I was assuming that for the ramification profiles $(3,2,2)$ and $(2,2,2,2)$, I could always apply a linear automorphism on the range to make the parts corresponding to $(2,2)$ in both be poles and zeros, and then apply a linear automorphism on the domain to bring the two divisors $2P + R - 2Q - S$ and $2P' + R' - 2Q' - S'$ into each other, and from that apply Riemann-Roch to limit the dimension of the space of permitted functions. Of course the problem now that I think through this more carefully is that I cannot interchange $2P + R - 2Q - S$ and $2P' + R' - 2Q' - S'$ freely in $P^1$.
