Why is it possible to derive the Lie algebra given its Lie group's constraint?

Question

I would like to intuitively understand how to derive the Lie algebra (the tangent space at the identity element $\mathbf{I}$?) of a Lie group $\mathcal{G}$ given the constraints that apply for the group.

After watching this excellent and easy-to-follow video, I decided to try to derive the basis vectors (are they called generators?) of the Lie algebra of the unit complex numbers, which can represent 2D rotations.

I chose this group only as an example for the sake of this question, so that I could describe my question on an easy group, which is easy to visualize in mind. However, my question is not limited specifically to this simple group.

The following is my current understanding

The unit complex numbers are a Lie algebra $\mathcal{M}$, whose manifold can be defined as $\mathcal{M} = \{\mathbf{z} \in \mathbb{R}^2 \mid {\mathbf{z}^2_x} + {\mathbf{z}^2_y} = 1 \}$. In my understanding, this constraint "filters-out" points from $\mathbb{R}^2$ and leaves just the points on the unit circle centered at the origin, creating the manifold. The tangent space at the point $\mathbf{I} = \begin{pmatrix}1\\0\end{pmatrix} \in \mathcal{M}$ is then called the Lie algebra $\mathfrak{m}$ of the Lie group $\mathcal{M}$.

(I opted to work in $\mathbb{R}^2$ instead of $\mathbb{C}$, so that is why I re-wrote the constraint as ${\mathbf{z}^2_x} + {\mathbf{z}^2_y} = 1$ instead of $\mathbf{z^*z} = 1$ for complex numbers (Where $\mathbf{z^*}$ is the complex conjugate.).)

By following the video mentioned above, I consider a random path on the manifold $X(t): \mathbb{R} \to \mathcal{M}$, which at point $t=0$ passes through the identity point $\mathbf{I}$.

Let us state what we know: $$ \begin{aligned} X(t) &= \begin{pmatrix}a(t)\\b(t)\end{pmatrix}\\ X(0) &= \begin{pmatrix}1\\0\end{pmatrix} \Longrightarrow a(0)=1, \,b(0)=0\\ X'(t) &= \begin{pmatrix}a'(t)\\b'(t)\end{pmatrix}\\ X'(0) &= \begin{pmatrix}p\\q\end{pmatrix} \in \mathfrak{m}\\ \end{aligned} $$

We want to derive the structure of the Lie algebra, which we will do by differentiating the group's constraint:

$$ \begin{aligned} {\mathbf{z}^2_x} + {\mathbf{z}^2_y} = 1\; &\Longleftrightarrow\; a(t)^2 + b(t)^2 = 1\\ \\ \frac{\partial}{\partial t}\left( a(t)^2 + b(t)^2 \right) &= \frac{\partial}{\partial t}\left(1\right)\\ 2a(t)a'(t) + 2b(t)b'(t) &= 0\\ \\ 2a(0)a'(0) + 2b(0)b'(0) &= 0\\ a'(0) &= 0\\ p &= 0 \end{aligned} $$

Now we know that the $p$ component is zero, which provides the constraint that defines the Lie algebra (i.e. the basis vector(s) of the tangent space at the point $\mathbf{I}$ of the manifold $M$):

$$ \begin{aligned} X'(0) &= \begin{pmatrix}0\\q\end{pmatrix} \in \mathfrak{m}\\ \mathfrak{m} &= \left\{\begin{pmatrix}p\\q\end{pmatrix} \mid p = 0\right\} = \text{span}\left\{ \begin{pmatrix}0\\1\end{pmatrix} \right\} \end{aligned} $$

This means the Lie algebra $\mathfrak{m}$ is spanned by just one basis (tangent) vector $e = \begin{pmatrix}0\\1\end{pmatrix}$.

Now that I showed how I derived the Lie algebra generator, let me ask my questions.

What is not clear to me

When you want to compute the tangent space of a function $f(x,y)$ at a point $p = \begin{pmatrix}a\\b\end{pmatrix}$, you compute the partial derivatives $\partial_x f(a, b)$ and $\partial_y f(a, b)$. Those partial derivatives are derivatives along the $X$ and $Y$ axis. However, when you have an implicit definition of a manifold like $RR^T = I$ for the $SO(3)$ Lie group or the above mentioned $\mathbf{z^*z} = 1$, there aren't parameters like $x,y,z$ w.r.t. which you could take the derivative.

1. Why do we have to consider a random path on the manifold $X(t)$? This is not something I would intuitively come up with. Where does this approach logically follow from?
2. Is construction of a random path on the manifold the only way of arriving at the basis vector(s)?
3. Why is $X(t)$ parametrized by only one parameter $t$, regardless of the dimension of the manifold? Why does it work? Why don't we choose e.g. 2 parameters $X(s,t)$?
4. What does it mean to take the derivative of the group's constraint (e.g. $\frac{\partial}{\partial t}\left( a(t)^2 + b(t)^2 \right) = \frac{\partial}{\partial t}\left(1\right)$). How can you visualize it? I do understand what does it mean when you take a derivative of a function w.r.t. its parameters, but I have no intuition nor visualization for taking the derivative of an equation.
5. If we consider a random path $X(t)$ through the identity point $I$, the path can have arbitrary orientation (it can "come into" the point $I$ from infinitely many directions). Aren't the basis vectors different for each particular path $X_n(t)$?
6. Is it guaranteed that all basis functions found in this way are orthonormal w.r.t. each others? Is it needed to normalize and orthogonalize them afterwards?