Do carmo: theorem of turning tangents --- notational confusion Theorem statement
Let $\mathbf x: U \subseteq \mathbb R^2 \to S \subseteq \mathbb R^3$ be a parametrization compatible with the orientation of $S$. Assume further that $U$ is homeomorphic to the open disk in the plane.
Let $\alpha: [0, l] \to \mathbf x(U) \subseteq S$ be a simple, closed, piecewise regular parametried curve with vertices $\alpha(t_i)$ and external angles $\theta_i, i = 0,\dots,k$.
Let $\phi: [t_i, t_{i+1}] \to R$ be differentiable functions which measure at each $t \in [t_i, t_{i+}]$ the positive angle from $\mathbf x_u$ to $\alpha'(t)$.
Theorem of turning tangents: With the above notation:
$$
\sum_{i=0}^k (\phi_i(t_{i+1}) - \phi_i(t_i)) + \sum_{i=0}^k \theta_i = \pm 2 \pi
$$
Questions

*

*What is the quantity $\phi_i$? I don't follow from the definition what $\mathbf x_u$, and what is it trying to capture by considering the angle between this $\mathbf x_u$ and $\alpha'(t)$?


*Where can I find a proof of this exact theorem? Do Carmo states this without proof. I would like to find a proof of this exact statement of the theorem --- I have found other proofs that invoke the gaussian curvature.
Picture for reference

 A: There is one element however that you overlooked, which will greatly simplify the intuition for the problem. Just one page earlier, in order to settle the question of an angle at the cusp, Do Carmo states :

Let the closed simple curve $\alpha$ be contained in the image of a conformal parametrization.

The fact that we are interested in conformality is reflected in the proofs of all the Gauss-Bonnet Theorems, by the fact that the local GB requires isothermal coordinates, and that in order to prove the Global GB, we construct our triangulation such that each triangle is contained in such a parametrization $-$ isothermal coordinates are synonimuous with conformality (at least localy).
Notice also, that The theorem of turning tangents is crucial to Do Carmo's proof of local GB.
Remember that conformal maps preserve angles (not necessarily length).
Now, with that out of the way, and you, hopefully convinced that we are dealing with simple, closed curves that are contained in a single, conformal parametrization, we can answer your questions :
Given $x:U \subset R^2 \rightarrow S$ and a path $\alpha$ with trace in S. By conformality, the angle $\varphi(t)$  between $\alpha'(t)$  and $x_u$ in $S$ (upstairs) is the same as in $U$ (downstairs) (by applying $x^{-1}(\alpha)$).
Therefore, let's just stay downstairs, since all the angles are preserved, $x_u$ downstairs is just $(1,0)$ and our path $\alpha(t)=x(u(t),v(t)$ becomes $(u(t),v(t))$. Therefore, $\varphi(t)$ is also just the angle between $(u'(t),v'(t))$ and $(1,0)$.
Look at the closed curve and let's make it so that there are no jumps (no $\theta_i$, our curve is differentiable everywhere).
Now, draw different closed curves, and look at how the angle between the tangent and $(1,0)$ changes. You will realise, that the tangent has to make one complete turn.
An even  simpler analogy is, point your arm in one direction (everywhere but straight up or down), and try to walk along some simple closed path, without changing the direction in which your arm is pointing, in the sense that your arm, at any given time, should remain parallel to where it was at any other. You will realise that at some point, you have to break your shoulder (when $\varphi$ reaches an angle of $\pi$ at the latest). In fact, you will have to make one complete turn around your arm (unhealthy). Thus, totaling an angle of $\pm 2 \pi$ along the way.
I hope this answers your first question.
If you understand this relationship betweeen the upstairs and downstairs, the proof of theorem 2, p.396, in this Version of Do Carmo should grant you satisfaction.
P.S. The newer editions of Do Carmo add "for plane curves" to the theorem of turning tangents.
