Let $S^1$ be a circle (i.e. a closed $1$-dim. manifold, without fixing metric, nor parallel transport). Let $x\in S^1$. Then it seems to me that there is no canonical projection $\pi : T_x S^1 \rightarrow S^1$.
And it also does not make sense to integrate a vector field $v$ on $S^1$ along a curve in $S^1$.
Is this correct?
On
If your manifold happens to be a compact Lie Group $G$ (as $S^1$) then the exponential map $$\exp\colon \mathfrak g \to G$$ is a surjective map from the tangent space at the identity to the Lie Group. By precomposition with the pushforward of left multiplication with $g^{-1}$ you get a surjective map $$exp∘(L_{g^{-1}})_{*}:T_gG \to G.$$ In the special case of $S^1$ and $g=1$ this map corresponds to the universal covering map, which is not true in general. Of course, the answer depends on what you exactly meant by projection in this case. In general, a compact Riemannian manifold $(M,g)$ suffices to construct a surjective smooth map from the tangent space $T_x M$ to the manifold. It is given by the Riemannian exponential map. The Hopf-Rinow theorem guarantees that this map is surjective when $M$ is compact.
For the canonical projection: there exists a canonical projection $\pi:TS_{1}\rightarrow S_{1}$ given by $(x,v)\mapsto x$. However, there is not an equally obvious projection $\pi:T_{x}{S_{1}}\rightarrow S_{1},$ because in fact $T_{x}{S_{1}}$ is diffeomorphic to the real line. There is a projection from $\mathbb{R}$ to $S_{1}$ given by $x\mapsto e^{ix},$ but this is a very special case.