How can a velocity vector be determined literally as a derivative in the calculus sense? I am following Tu's book on manifolds. On pages 178-179 he proves that the Lie algebra of $\mathrm{SL}(n,\mathbb{R})$ is the set of traceless real $n\times n$ matrices. Specifically he writes:

Suppose $X \in T_I \mathrm{SL}(2,\mathbb{R})$. There is a curve $\gamma$ and $\varepsilon > 0$ such that $\gamma : (-\varepsilon,\varepsilon) \to \mathrm{SL}(2,\mathbb{R})$, $\gamma(0) = I$ and $\gamma'(0) = X$. Being in $\mathrm{SL}(2, \mathbb{R})$, this curve satisfies $\mathrm{det} (\gamma(t)) = 1$.

Then comes the part I cannot understand, he then says:

Now differentiate both sides with respect to $t$ and evaluate at $t=0$. On the right hand side we have $0 = \frac{d}{dt}\big|_{t=0} 1$, and on the left hand side we have $$ \frac{d}{dt}\big|_{t=0} \mathrm{det} (\gamma(t)) = (\mathrm{det \circ \gamma})_* \left( \frac{d}{dt}\big|_{0} \right) \qquad (\dagger) $$ [...]

As far as I can tell, in $(\dagger)$ he makes the literal interpretation that the tangent vector $(\mathrm{det} \circ \gamma)_{*0} \frac{d}{dt}\big|_{0}$ is the calculus derivative: $$D_{\gamma(t)} \mathrm{det} \cdot D_t \gamma $$ I cannot understand this because a calculus derivative is a real number, while the tangent vector is a derivation. Could you please explain in a pedagogical way what is going on?
 A: Edit: I've restructured this quite a bit, but I think it gets at the core issue more effectively now.
Given a differential map between manifolds $f:M \to N$, at each $p \in M$ we have a push forward map $f_*:T_p M \to T_{f(p)}N$ which is defined by the following: if $v \in T_p M$ and the path $\sigma: (-\epsilon,\epsilon) \to M$ satisfies $\sigma(0)=p$, $\sigma'(0)=v$, then $f_*(v) = (f \circ \sigma)'(0) \in T_{f(p)}$. Oftentimes, one writes this as 
$$f_*(v) = \ \frac{d}{dt} \Bigr |_{t=0} (f \circ \sigma)(t) \qquad (*)$$
in place of the prime notation. Of course, this latter notation does not indicate differentiation in the calculus sense explicitly, which would not make sense on a general manifold-- it is a symbol representing the tangent vector to the path $(f \circ \sigma)(t)$ at $t=0$.
Now, in your case, with the map $f=\det \circ \gamma : \mathbb{R} \to \mathbb{R}$, your equation $(\dagger)$ is precisely the equation $(*)$, where $v = \frac{d}{dt} \Bigr |_0$ and $\sigma(t)=t \in \mathbb{R}$ (note this indeed satisfies the defining property of $\sigma$). The point of confusion is that the RHS now has two potential interpretations: one in the formal sense as above, and one in the sense of the usual calculus derivative, and these seem to be at odds. So, the question really comes down to why does the differentiation notation in equation $(*)$ make sense? The crux of the issue is that the first interpretation represents an element of $T_{\det(\gamma(0))} \mathbb{R}$, while the second represents simply an element of $\mathbb{R}$.
To resolve this, recall that for any $p \in \mathbb{R}^n$ we have a canonical identification of $T_p \mathbb{R}^n$ with $\mathbb{R}^n$ via 
$$\sum_{i=1}^n x_i e_i \mapsto \sum_{i=1}^n x_i \frac{\partial}{\partial x_i}\biggr| _p$$
Where $\{e_i\}$ are the standard basis vectors in $\mathbb{R}^n$ (i.e., the $n$-tuples with $n-1$ zeros and $1$ one in the $i$th spot) and $\{x_i\}$ the associated coordinates. In this way, we often think of $T_p \mathbb{R}^n$ as 
$\mathbb{R}^n$ itself and use them interchangeably. The resolution to the above confusion is achieved by making this identification in the case $n=1$, as the two interpretations of $(*)$ are then identical. More generally, this is the case whenever the RHS side of $(*)$ can be interpreted as a usual calculus derivative.
A: Although the expression $f(t) = \text{det}(\gamma(t))$ is a good old 1-variable calculus function, and so its derivative is indeed obtained from the formula
$$f'(0) = \lim_{h \to 0} \frac{1}{h}(f(0+h) - f(0))
$$
the problem is that the composition $f = \text{det} \circ \gamma$ is not a good old 1-variable composition. 
Instead, $\gamma(t)$ is a function of the form $\gamma : \mathbb R \to \mathbb R^{n^2}$, and $\det$ is a function of the form $\det : \mathbb R^{n^2} \to \mathbb R$. So, in calculating $f'(t)$ you cannot use the 1-variable chain rule, you have to use the multivariable chain rule, which we can think of as matrix multiplication
$$f'(t) = D(\text{det}) \cdot D\gamma
$$
where $D(\text{det})$ is the $1 \times n^2$ row vector of partial derivatives of the determinant formula (i.e. the partial derivatives of "Kramer's Rule"), and $D\gamma$ is the $n^2 \times 1$ column vector of derivatives of the entries of $\gamma$.
Can you work it out from here?
