Let's look at your second example. Let $p(z, w) = w^2 - z^3 + z^2 + z$, and let $Y = \{ p(z, w) = 0 \}$. Then,
$$p(z, i) = -z^3 + z^2 + z - 1 = -(z - 1)^2 (z+1)$$
so I claim $(1, i)$ has ramification index $2$ while $(-1, i)$ has ramification index $1$. Indeed, observe that
\begin{align}
\frac{\partial p}{\partial z} & = -3 z^2 + 2 z + 1 \\
\frac{\partial p}{\partial w} & = 2 w
\end{align}
so by an inverse function theorem argument, we find that $(z, w) \mapsto z$ is locally a chart near both $(-1, i)$ and $(1, i)$. In this chart, your function $g : Y \to \mathbb{C}$ is given by $z \mapsto \sqrt{z^3 - z^2 - z}$. Let us take Taylor expansions around $\pm 1$:
\begin{align}
g(z) - i & = -i (z-1)^2 + O((z-1)^3) \\
g(z) - i & = -2 i (z+1) + O((z+1)^2)
\end{align}
Hence, the ramification index at $(1, i)$ is indeed $2$ and at $(-1, i)$ it is $1$.
Morally, what is going on is that your curves are dense open subsets of projective curves. Indeed, your first curve is given in homogeneous coordinates by
$$w^3 - z (z^2 - u^2) = 0$$
and your second curve is given by
$$w^2 u - z^3 + z^2 u + z u^2 = 0$$
and one can check by hand that these curves are smooth "at infinity", so we have the desired embedding of the original affine algebraic curves into projective (hence compact) algebraic curves. Degree is well-defined on the latter, so is well-defined on the former by restriction; the only trouble is that there may be "missing" preimages and so the equation relating degrees and ramification indices becomes an inequality:
$$\text{deg}(g) \ge \sum_{x \in g^{-1} \{w\}} \nu_x (g)$$
For example, take the affine hyperbola $z w - 1 = 0$ and the projection $(z, w) \mapsto w$; this function has degree $1$ (once we embed it in the projective closure), but obviously there are no preimages of $0$ in the affine hyperbola.
Let's develop a generic method of dealing with affine plane curves. Let $p : \mathbb{C}^2 \to \mathbb{C}$ be a polynomial function in two variables, and suppose $Y = \{ p(z, w) = 0 \}$ is a smooth algebraic curve. Let $f : Y \to \mathbb{C}$ be the projection $(z, w) \mapsto w$. For each fixed complex number $b$, we get a polynomial function $p(-, b)$, say of degree $d$. Now, because $\mathbb{C}$ is algebraically closed, we can write
$$p(z, b) = c (z - a_1)^{e_1} \cdots (z - a_n)^{e_n}$$
for some distinct complex numbers $a_1, \ldots, a_n$, $c \ne 0$, and positive integers $e_1, \ldots, e_n$, such that $e_1 + \cdots + e_n = d$. Suppose also that
$$\frac{\partial p}{\partial w}(a_i, b) \ne 0$$
for all $a_i$; then an inverse function theorem argument shows that $(z, w) \mapsto z$ is a chart near each $(a_i, b)$. I claim that the ramification index of $f$ at $(a_i, b)$ is $e_i$ under these hypotheses. Indeed, when $z$ is a local parameter, we have
$$0 = \frac{\partial p}{\partial z} + \frac{\mathrm{d} w}{\mathrm{d} z} \frac{\partial p }{\partial w}$$
so if $e_i > 1$, we have $\frac{\partial p}{\partial z} (a_i, b) = 0$, so we must have $\frac{\mathrm{d} w}{\mathrm{d} z} (a_i) = 0$ because $\frac{\partial p}{\partial w} (a_i, b) \ne 0 $ by hypothesis – implying $f(z) - b = O((z - a_i)^2)$. Playing around with total derivatives more, we eventually find that the first non-zero coefficient of $f(z) - b$ around $a_i$ is the coefficient of $(z - a_i)^{e_i}$, as required.
On the other hand, when we have $\frac{\partial p}{\partial w} (a_i, b) = 0$, then by non-degeneracy we must have $\frac{\partial p}{\partial z} (a_i, b) \ne 0$, and we must have $e_i = 1$ and $(z, w) \mapsto w$ is a chart near $(a_i, b)$. But then obviously the ramification index of $f$ at $(a_i, b)$ must be $1$. So in either case the ramification index of $f$ at $(a_i, b)$ is equal to $e_i$. Convenient, no?
When you discuss ramification points of $\pi$, the equation should be $x^d + y^d = 0$, not $x^d + y^d = 1$.
Also, whenever you wrote about $d$th roots of unity, you needed the $d$th roots of $-1$.
The rest seems right. Yes, the number of preimages of the ramification points is all that you need in order to apply Hurwitz's formula, it is not necessary to find the degree of each preimage.
Best Answer
I don't think you should be taking the affine chart $Z\ne 0$ in this case. You have a map $C\rightarrow \mathbb{P}^1$ and if you want to understand this map affine locally, you should take an affine chart on $\mathbb{P}^1$ and consider its preimage instead.
Let's call homogeneous coordinates on $\mathbb{P}^2$ by $X,Y,Z$ and homogeneous coordinates on $\mathbb{P}^1$ to be $U,V$. Then the pre-image of $D(U) =\{ [U:V]: U\ne 0\} \cong \mathbb{A}^1$ is $D(X)\cap C = \{[X:Y:Z]\in C: X\ne 0 \}$. In particular the corresponding affine curve is $z^2y^2 = 1 + y^4 + z^4$, and in this case your map is given by projection to $y$-coordinate. To determine the branch points then it suffices to determine all $y$ such that there are less than 4 distinct roots for $z$, which now becomes a problem in solving quadratic equations.
Then you can also look at the other chart $D(V)$ and the corresponding affine curve and determine if $\infty$ (which corresponds to the point $[0:1]$) is a branch point by the same argument.