I want to check if
$M := \{x \in \mathbb{R}^2 | \exists t>0: x_1 = sin(t) /t, x_2 = cos(t) /t \} $
is a $C^1$-manifold.
To do this I would need to find for every point $x \in M$ a $C^1$-function $\phi : U \rightarrow ?$ with the known properties, but I do not have an idea for such a function, and if there should not exist one, I also do not see an argument which would show this.
Let's rewrite your set: $$ M = \{ (\frac{sin t}{t}, \frac{\cos t}{t}) \mid t \in \mathbb R^{+}\}. $$
As $t$ grows large, you've basically got something that's like a spiral; when $t$ gets near $0$, you've got something that looks like the tail end of a snail, where the "tail" extends out to infinity on the $x$-axis.
Let's divide the reals into two categories: those for which $$ u(t) = (\frac{sin t}{t}, \frac{\cos t}{t}), $$ expressed in polar coordinates, has an angular coordinate that is within $\pi/4$ of a multiple of $\pi$, which I'll call $A$, and those for which it's not (so that it's within $\pi/4$ of a multiple of $\pi$ plus $\pi/2$), which I'll call $B$. The ones that are in both sets...put them in whichever set you like.
For point $t = a$ in group $A$, consider the map, defined on the interval $I_a$ consisting of points $t$ with $\min(a - .01, 0) < t < a + .01$, that takes $(x, y)$ to $y$.
This map, when restricted to the image $u(I_a)$, is $C^1$ and has all the properties you need.
For those in group $B$, use the map $(x, y) \mapsto x$ instead.
More generally: For most manifolds in euclidean space, projection onto one coordinate plane or another serves as a very good coordinate chart; in fact, the implicit function theorem guarantees this.