Partial Derivatives and Linear Map Proposition:
Suppose that $T : R^n \to R^m$
is the linear transformation defined by
$T(x) = Mx$
for some m × n matrix $M$. Then
$DT(x) = M$
for all points $x \in R^n$
where $D$ is the partial derivative matrix. (Jacobian?)

Question:
I don't understand what is being said. $T(x)$ is a linear transformation on $x$. How does the partial derivative of $T(x)$ lead to the transformation matrix. Neither do I have an algebraic intuition nor a geometric one.

Further,
How is the total derivative of $g(x,y,z)$ equal to $ Dg(x,y,z) \begin{pmatrix}
x \\
y\\
z \end{pmatrix}$? 
This is stated without proof. There is a chance, I made a wrong interpretation so I am pasting the portion of the text where it appears.

Is it that the change in $x$ in all dimensions of the output of $g(x,y,z)$ multiplied by $x$ and similar for y and z gives a total derivative. I don't seem to understand.I know the total derivative is a derivative taking into account that other variables are not constant during differentiation by one variable.
 A: For algebraic intuition:  The derivative of a function is the best linear approximation to it.  In the case when the function is linear already, it's its own best linear approximation, hence it's its own derivative.
As a reminder, for multivariate $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$, the derivative $Df(x)$ at the point $x\in \mathbb{R}^m$ is defined to be a linear map $A:\mathbb{R}^m\rightarrow\mathbb{R}^n$ satisfying
$$\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x) - Ah}{\|h\|}=0$$
It's not too hard to show that if $A$ exists, it's unique.
Now, we simply check that for $T(x) = Mx$, everything works.
$$\lim_{h\rightarrow 0} \dfrac{T(x+h)-T(x) - Mh}{\|h\|} = \lim_{h\rightarrow 0} \dfrac{M(x+h)-Mx-Mh}{\|h\|} = \lim_{h\rightarrow 0}\dfrac{Mx+Mh-Mx-Mh}{\|h\|}=0$$ since the numerator is $0$.
Edit to address the comment
In answer to the comment, the derivative at a point is the best linear approximation at that point (though we think of the point as being shifted to the origin).  Thus, in your example of $f(x) = x^3$, the derivative, $3x^2$, at a point (say $x=2$) gives a slope of $12$.  The linear approximation this corresponds to is $12x$.
In fact, this issue is one it took me a long time to come to terms with.  For functions $f:\mathbb{R}\rightarrow\mathbb{R}$, we think of the derivative of the whole function as a new function $f'(x)$.  With more variables, we change view point:  The derivative of a function at a point is itself a function (linear transformation).  The relationship between the 2 is that a linear transformation from $\mathbb{R}\rightarrow\mathbb{R}$ can be identified with a single real number $\mathbb{R}$ via a pretty canonical choice of basis.  The derivative of a function $f:\mathbb{R}\rightarrow\mathbb{R}$ at a point $p$ is the linear map "multiply by $f'(p)$".
