Why is the algebraic value of the covariant derivative equal to $\langle dw/dt, N \wedge w\rangle$? 
Why is the algebraic value of the covariant derivative equal to
  $\langle dw/dt, N \wedge w\rangle$?

Just a simple statement made by do Carmo. I suppose my linear algebra is rusty because I can't see how it's true. Although I understand why $Dw/dt=[Dw/dt](N \wedge w)$.
Page 248 of do Carmo states $Dw/dt=\lambda(N \wedge w(t))$ which I understand since $Dw/dt$ is an orthogonal projection. He then defines $\lambda$ as a real number denoted by $[Dw/dt]$ as the algebraic value of the covariant derivative of $w$ at $t$. Then says "Observe that the sign of $[Dw/dt]$ depends on the orientation of $S$ and that $[Dw/dt]=\langle dw/dt, N \wedge w\rangle$ where $w$ is a differentiable field of unit vectors along a parametrized curve". 
 A: Since $N$ and $w$ are unit vectors, so is $N \times w$.  Then
$$\left\langle \frac{Dw}{dt}, N \times w \right\rangle = \left[ \frac{Dw}{dt} \right].$$
Since $N\times w$ is a tangent vector, being orthogonal to $N$, its inner product with anything in the normal direction is $0$.  $Dw/dt$ is the tangential component of $dw/dt$ and so the two vectors differ by a multiple of $N$, say $qN$.  That is, 
$$\frac{dw}{dt} = \frac{Dw}{dt} + qN$$
for some function $q$.  Then we have
$$\left\langle \frac{dw}{dt}, N \times w \right\rangle = \left\langle \frac{Dw}{dt}, N\times w \right\rangle + \left\langle qN, N \times w \right\rangle = \left\langle \frac{Dw}{dt}, N\times w \right\rangle = \left[ \frac{Dw}{dt} \right].$$
A: I will first show why $\dfrac{\mathrm Dw}{\mathrm dt}(t)=\lambda(t)(N\wedge w).$ From the derivation, one can obtain the required equality as well.
Since $w(t)$ is a unit vector field, it is easy to see that $\dfrac{\mathrm dw}{\mathrm dt}(t)$ is normal to $w(t)$. The set of all vectors perpendicular to $w(t)$ forms a subspace of dimension $2$, say $w(t)^{\perp}$, in $\mathbb R^3$. As $N$ and $N\wedge w$ are linearly independent elements in $w(t)^{\perp}$, they form a basis for the same. So $$\dfrac{\mathrm dw}{\mathrm dt}(t)=\mu(t)N+\lambda(t)(N\wedge w),$$for some functions $\mu(t)$ and $\lambda(t)$. Thus $$\dfrac{\mathrm Dw}{\mathrm dt}(t)=\dfrac{\mathrm dw}{\mathrm dt}(t)-\left\langle\dfrac{\mathrm dw}{\mathrm dt}(t), N\right\rangle N=\lambda(t)(N\wedge w).$$

Note that $\left\langle\dfrac{\mathrm dw}{\mathrm dt}(t), N\wedge w\right\rangle=\lambda(t)$ follows easily from the last step.
