I‘m first going to comment on your two questions and then, for completeness, I‘ll give a sketch of proof for the exercise.
1) You are right that if we look at the curve as $t\to \gamma(e^{2\pi i t})$ then there is a parallel vector field $V(t)$ along $\gamma$ with $V(0)=v$ for every $v\in T_pM.$ However it might be that $V(1)\neq V(0).$ But as now we see the curve as a mapping $\gamma:S^1\to M$ we also want a parallel VF $V$ as a map from $S^1$ sucht that $V(t)\in T_{\gamma(t)}M$ and therefore the parallel VF should have the same value when starting and when „finishing the whole round“.
If this is not quite clear, then don‘t worry: In my answer below you can still think of the curve as a map $\gamma:[0,1]\to M,$ but we will see that there is a parallel vector field with $V(1)=V(0).$
2) Let $\omega$ be an orientation form for $M,$ i.e. a non-vanishing $n-$form (where $n$ is the dimension of $M$).
For each $t\in [0,1]$ we can define an orientation on the orthogonal complement of $\gamma‘(t)$ by declaring $v_1,...,v_{n-1}$ to be positively oriented if
\begin{equation}
\omega_{\gamma (t)}(v_1,...,v_{n-1},\gamma'(t))>0.
\end{equation}
Let $A:\left(\mathbb{R}\gamma'(0)\right)^{\perp}\to \left(\mathbb{R}\gamma'(0)\right)^{\perp}$ be the restriction of the parallel transport map along $\gamma$ to the orthogonal complement of $\gamma'(0)=\gamma'(1).$
$A$ is clearly an isometry and also it is orientation preserving. To see the latter just observe that when $e_1,...,e_{n-1}$ is a positively oriented ONB of the orthogonal complement of $\gamma'(0)$ and if $E_1(t),...,E_{n-1}(t)$ denote their parallel transports along $\gamma,$ then for all $t$ we have
\begin{equation}
\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))\neq 0
\end{equation} because $E_1(t),...,E_{n-1}(t),\gamma'(t)$ is a basis for $T_{\gamma(t)}M$ and therefore, as $e_1,...,e_{n-1}$ is positively oriented,
\begin{equation}
\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))>0.
\end{equation} So $E_1(1),...,E_{n-1}(1)$ has the same orientation as $e_1,...,e_{n-1}.$ Because of $E_i(1)=A(e_i),$ $A$ is orientation preserving.
Because $n$ is even, Lemma 3.8 in Do Carmo impliess that there is a $v\in T_{\gamma(0)}M$ such that
\begin{equation}
V(1)=A(v)=v,
\end{equation}where $V(t)$ denotes the parallel transport of $v$ along $\gamma.$
Now to the rest of the exercise:
Let $v$ and $V(t)$ such as we obtained it in 2).
Then
\begin{equation}
h(s,t)=exp_{\gamma(t)}(sV(t))
\end{equation} is a variation of $\gamma.$ By the same calculation as in the proof of the Synge-Weinstein-Theorem we get
\begin{equation}
\frac{1}{2}E''(0)=-\int_0^1K(V(t),\gamma'(t))dt<0
\end{equation}
and so there is a $s_0$ such that the closed curve $c(t):=h(s_0,t)$ satisfies $l(c)<l(\gamma)$ (details again at the end of the proof of the Synge-Weinstein-Theorem).
One cannot trust Wikipedia too much on these matters: Anybody can edit a Wikipedia article. The Wikipedia article is sloppy about distinguishing compact and noncompact surfaces. The Uniformization Theorem has the following equivalent forms:
A. Every simply-connected Riemann surface is biholomorphic to $S^2$ or to ${\mathbb C}$ or to the unit disk $\Delta$ in ${\mathbb C}$.
B. Every connected Riemann surface $X$ is biholomorphic either to $S^2= {\mathbb C}P^1$ (with its standard complex structure), or to the quotient of $U={\mathbb C}$ or of $U=\Delta$ by a group $\Gamma$ of linear-fractional transformations of $U$ acting on $U$ freely and properly discontinuously.
C. Every connected Riemannian surface $(S,g)$ admits a positive smooth function $\lambda$ such that $(S, \lambda g)$ is a complete Riemannian manifold of constant curvature. (Note that the surface $S$ is not required to be oriented.)
Remark. i. In all three formulations, the surface $S$ is not required to be compact. The compactness assumption made by the Wikipedia article is totally unnecessary. Compactness is used in some of the proofs but not in other proofs.
ii. There is no way to prove the UT for general noncompact surfaces from the UT for compact surfaces.
Equivalence of the three statements (A, B and C) is not hard to establish. The key facts are the existence of a conformal Riemannian metric on every Riemann surface and that every group $\Gamma$ as in Part B, acts on $U$ isometrically with respect to the Euclidean or hyperbolic metric respectively.
Regarding the Ricci flow, what is true is that there is a proof of the Uniformization Theorem (UT) for compact (closed) Riemann surfaces via the Ricci Flow (RF). This result by itself does not imply the full UT.
In order to prove the UT for noncompact surfaces using RF one would have to work much harder than in the compact case and I am unaware of such a proof in the literature. Even short-term existence of the flow becomes a problem. See for instance
Xiaorui Zhu, Ricci Flow on Open Surface, J. Math. Sci. Univ. Tokyo 20 (2013), 435–444.
for some partial results on proving UT via RF on open surfaces.
Of course, if your (say, simply connected) Riemann surface is compact, then RF indeed does the job: First equip your surface $X$ with an arbitrary conformal Riemannian metric $g_0$ (i.e. a metric which in the local holomorphic coordinates of $X$ has the form $\rho_k(z)|dz|^2$). For a proof of existence of such a metric see my answer here.
Then apply the normalized RF to $g_0$. This, in finite time, converges to a constant curvature metric $g_T$ (the curvature has to be positive). For surfaces, RF preserves the conformal class of the metric. Hence, $g_T$ is still a conformal Riemannian metric on $X$. After rescaling, $g_T$ has curvature $1$. Now, use the theorem (due to Killing and Hopf in all dimensions) that all compact simply-connected surfaces of curvature $1$ are isometric to each other. Hence, $(X,g_T)$ is isometric to the standard unit sphere $S^2$. The isometry $f: X\to S^2$ has to be conformal in the sense of Riemannian geometry, hence (after changing orientation if needed) is conformal in the sense of complex analysis. qed
Best Answer
Based on the newest edit, I've made some adaptations.
First, if $M$ is not compact, the result is not true. As said in the comments, the punctured plane with the induced metric is an example.
If $M$ is compact, the result is true. In fact, we have that the following is true:
Those loops are necessarily geodesics, by variational reasons.
One way to see why this is true is via a Morse-theory approach, with a full proof being given in Klingenberg - Lectures on Closed Geodesics. The gist of it is to pick a loop close to the infimum on the class and flow it under (minus) the gradient flow of the energy functional in the free loop space (this has an obvious analogy with the finite-dimensional case when we flow via gradient and converge to a critical point).
To see how the result we mentioned implies what you want, just notice that a variation of a geodesic stays in the same class. Thus, if the geodesic minimizes length on the class, it is a local minimizer.
EDIT: The book by Klingenberg gives a reference which does not use the machinery behind the proof I alluded to above. He says:
He attributes the proof (in special cases) to Hadamard and Cartan (although he says that the later has an error), and says that Berger - Lectures on geodesics in Riemannian geometry has a full proof following this idea (I have not seen it).