Tangent space as the set of all derivations I am trying to get a grip on the concept of derivations at a point on a manifold by working out some concrete examples.

Let $M$ be a smooth manifold with or without boundary, and let $p \in M$. A linear map $v : C^{\infty}(M) \to \mathbb{R}$ is called a derivation at $p$ if it satisfies $$v(fg) = f(p)v(g) + g(p)v(f)$$
The set of all derivations of $C^{\infty}(M)$ at $p$ is denoted by $T_pM$ and is a vector space called the tangent space to $M$ at $p$.

Now let's look at an example; from single-variable calculus.
Take $M = \mathbb{R}$ and $f(x) =\sin(x)$ and $p = \pi$. We have $f'(p) = \cos(p) = -1$.
Now let's look at the general definition above, and note that for any $a \in \mathbb{R}^n$, the $n$ derivations $$\frac{\partial}{\partial x^i}|_{a} \ \text{ defined by } \ \frac{\partial}{\partial x^i}|_{a}f = \frac{\partial f}{\partial x^i}(a)$$
for $i \in \{1, .., n\} $ form a basis for $T_a(\mathbb{R}^n)$ and has dimension $n$.
So pick $v \in T_p(\mathbb{R}^1)$, since $v$ is a linear-combination of basis elements in a vector space of dimension $1$, we have $v = c \cdot \frac{\partial}{\partial x}|_{p}$ for some $c \in \mathbb{R}$. So $v(f)  = c \cdot \frac{\partial f}{\partial x}|_{p} = c\cdot \frac{d f}{d x}|_{p} = c\cdot cos(\pi) = -c$ for some $c \in \mathbb{R}$
But we need $v(f) = -1$, for the computation from the general definition and usual calculus to coincide. Have I done anything wrong here? Shouldn't $v(f) = -1$?

You have effectively done everything correctly.
You recognised that $\frac{\partial}{\partial x}|_{p}$ was a basis for the tangent space at $p$ and $\frac{\partial f}{\partial x}|_{p} = -1$
An arbitrary tangent vector at $p$ is thus $v = c\frac{\partial}{\partial x}|_{p}$ for some $c$ so by linearity $v(f)= c\frac{\partial f}{\partial x}|_{p} = c(-1) = -c $ which is what you showed.
In other words, you came to the correct conclusion.
