Determine the tangent vector at any point $p=(x^1, x^2)$ on the circle $(x^1)^2 +(x^2)^2 = a^2$
The following problem above and its solution below was given in my class.
Solution: Parametric equations for the circle are $x^1 = a\cos t$ and $x^2 = a \sin t$. The tangent vector $v$ is given by $$v_p = v^1e_1 +v^2e_2$$ where $e_1 = \frac{\partial}{\partial x^1}$ and $e_1 = \frac{\partial}{\partial x^2}$. Note that $$v^1 = \frac{dx^1}{dt} = -x^2$$ and $$v^2 = \frac{dx^2}{dt} = x^1$$ and so we arrive at $$v_p = -x^2 \frac{\partial}{\partial x^1} +x^1 \frac{\partial}{\partial x^2}$$
Now I want to make the above solution as rigorous as possible.
Now the way I understand it (using the book Introduction to Smooth Manifolds by John Lee as a reference), by tangent vector my lecturer means geometric tangent vector which is the element $v_p :=(p, v) \in \mathbb{R}^n_p = \{p\} \times \mathbb{R}^n$.
Now $$T_p(\mathbb{R}^n) = \left\{ w : C^{\infty}(\mathbb{R}^n) \to \mathbb{R} \ | \ w \text{ is linear and } w(f \cdot g)= f(a)\cdot w(g) + g(a)\cdot w(f)\right\}$$ so in words $T_p(\mathbb{R}^n)$ is the set of all smooth linear maps from $\mathbb{R}^n$ to $\mathbb{R}$ satsifying the product rule above.
Some quick facts. For any $v_p \in \mathbb{R}^n_p$, the map $D_v|_p : C^{\infty}(\mathbb{R}^n) \to \mathbb{R}$ defined by $$D_v|_p (f) = \sum_{i=1}^n v^i \frac{\partial f}{\partial x^i} (p)$$ is a derivation. So basically for any point in $\mathbb{R}^n_p$, the map which takes the directional derivative of a smooth function on $\mathbb{R}^n$ in the direction $v$ at $p$ is a derivation.
Furthermore $T_p(\mathbb{R}^n)$ as basis $$\mathcal{B} = \left\{\frac{\partial}{\partial x^1}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right\} := \left\{D_{e_1}\big|_p, \dots, D_{e_n}\big|_p\right\}$$ where $e_i = (0, \dots, \underbrace{1}_{i^{\text{th}} \text{ position}} , \dots, 0)$
Moreover the map $\psi : \mathbb{R}^n_p \to T_p(\mathbb{R}^n)$ defined by $$\psi(v_p) = D_v|_p$$ is a vector space isomorphism. So $\mathbb{R}^n_p$ is isomorphic to $T_p(\mathbb{R}^n)$ as vector spaces.
Now with all of the above established, I'm trying to rephrase the above solution more formally.
I understand (please correct me if I'm wring) that the 'tangent vector' $v_p =(p, v)$ where $v = (v^1, v^2)$ is being identified with its image $\psi(v_p) = v^1\frac{\partial}{\partial x^1} + v^2 \frac{\partial}{\partial x^2}$
Now what I don't understand is how we determine $v^1$ and $v^2$ and why we need parametric equations for the circle. I don't see why $v^1$ and $v^2$ are calculated in the way they are.