Why does the term $A\mathbf{h}/\|\mathbf{h}\|$ appear in the definition of the derivative? My analysis course provides the following definition of the derivative of $f$, $Df(\mathbf{a})$:

Definition: A function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is differentiable at $\mathbf{a} \in \mathbb{R}^{n}$ if there exists a linear function $\mathbf{x} \mapsto A \mathbf{x}$ (where $A$ is an $m \times n$ matrix) from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$ such that
$$
\frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-A \mathbf{h}}{\|\mathbf{h}\|} \rightarrow \mathbf{0}.
$$

Now, if we unpack this and go back to the limit definition we get that $Df(\mathbf{a})$ is
$$
\lim_{\|\mathbf{h}\| \rightarrow  0 } \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})}{\|\mathbf{h}\|} - \frac{A \mathbf{h}}{\|\mathbf{h}\|}.
$$
What I don't understand is the term $\frac{A \mathbf{h}}{\|\mathbf{h}\|}$. Why is the product $A \mathbf{h}$ scaled by the norm of $\|\mathbf{h}\|$ as well?
Would appreciate further insight into this definition.
Additionally, based on this definition of the derivative, it is possible for directional derivatives to exist even the function is not differentiable at a particular vector $\mathbf{a}$. Why is this the case ?
Defining the directional derivative as:
$$D_{\mathbf{v}} f(\mathbf{a})=\lim _{h \rightarrow 0} \frac{f(\mathbf{a}+h \mathbf{v})-f(\mathbf{a})}{h}.$$
 A: First off, minor nitpick:  $A \mathbf{h}$ is not a dot product.  The dot product takes two $n$-dimensional real (or complex) vectors as input, and gives a real number as output.  $A$ is a matrix (not a vector) which represents a linear transformation $\mathbb{R}^m \to \mathbb{R}^n$ which acts on $\mathbf{h}$ (a vector in $\mathbb{R}^m$) in order to produce a vector in $\mathbb{R}^n$.
The Derivative
Addressing the question:  in one dimension, the derivative of $f$ at a point $a$ is generally defined as the limit of some kind of difference quotient, e.g.
$$ f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h},$$
assuming that this limit exists.  This definition is usually justified geometrically:  we think about approximating a tangent line to a curve via secant lines defined by points which are closer and closer to each other on the curve.  In that sense, the definition of the derivative in one dimension is natural.
Manipulate this identity to get
\begin{align}
0
&= \left( \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}\right) - f'(a) \\
&= \lim_{h\to 0} \left( \frac{f(a+h) - f(a)}{h} - f'(a) \right) && (\text{$f'(a)$ is a constant)} \\
&= \lim_{h\to 0} \frac{ f(a+h) - f(a) - hf'(a)}{h}.
\end{align}
With a little more manipulation, the definition of the derivative can be rewritten as

Definition: A function $f : \mathbb{R} \to \mathbb{R}$ is differentiable at a point $a \in \mathbb{R}$ if there exists a linear function $x \mapsto Ax$ from $\mathbb{R}$ to $\mathbb{R}$ (where $A$ is a $1\times 1$ matrix) such that
$$ \frac{f(a+h) - f(a) - Ah}{|h|} \to 0. $$

This version of the definition follows from the "usual" geometric definition, but encapsulates a slightly different idea.  In this case, we think of the derivative as the linear operator which best approximates how the function is changed by small perturbations around $a$.  That is, if we perturb $a$ by $h$, what will be the corresponding change in $f$, assuming that $f$ is (at least locally) linear?
The actual perturbation is the difference between $f(a+h)$ and $f(a)$, while the linear approximation of the perturbation is $Ah$.  The derivative $A$ is the best linear approximation in the sense that the relative error between actual perturbation and the linear approximation can be made as small as possible.
In higher dimensions, the same idea applies: if we perturb $\mathbf{x}$ by just a little, this is going to perturb $f$ by just a little.  We then want to find the linear function which best approximates this perturbation.  We assess this by looking at the relative error, i.e the difference between the actually perturbation and the linear approximation, scaled by the size of the perturbation in $\mathbf{x}$.  The result is the definition of the derivative given in the question.
Directional Derivatives
Regarding the existence of directional derivatives even if the derivative does not exist:  a directional derivative is the best linear approximation of a function in a particular direction, whereas the derivative is the best approximation in any direction.  It is possible for a function to be approximable in one direction, but not in every direction.  As a simple example, consider
$$ f(x,y) = |x|. $$
If you are moving parallel to the $y$-axis, the function is constant, so
$$ D_{\langle 0, 1\rangle} f(x,y) = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h} = \lim_{h\to 0} \frac{ |x| - |x| }{h} = 0. $$
Here, the directional derivative in the direction of $\langle 0, 1 \rangle$ exists (and is zero).  However, in any other direction, the derivative does not exist wherever $x=0$.  For example
$$ D_{\langle 1,0 \rangle} f(0,0) = \lim_{h \to 0} \frac{ f(0+h, 0) - f(0,0) }{ h }
= \lim_{h \to 0} \frac{ |0+h| - |0| } {h} = \lim_{h\to 0} \frac{|h|}{h}, $$
which does not exist.
A: Let's compare this to the ordinary definition of the derivative in 1 variable. Consider a function $f : \mathbb R \to \mathbb R$ and $a \in \mathbb R$. To say that $f$ is differentiable at $a$ means that the following limit exists:
$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}
$$
This limit is usually given its own notation:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}
$$
Now let's manipulate this around a bit:
$$\left( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \right) - f'(a) = 0
$$
Holding $a$ fixed, as $h$ varies we can treat $f'(a)$ as a constant, and then use the formula for limit of a constant:
$$\left( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \right) - \lim_{h \to 0} f'(a) = 0
$$
and now we can use the rule for limit of a difference:
$$\lim_{h \to 0} \left( \frac{f(a+h)-f(a)}{h} - f'(a) \right) = 0
$$
Combining into one fraction we obtain
$$\lim_{h \to 0} \frac{f(a+h)-f(a) - f'(a) h}{h} = 0
$$
I'm sure you can see that this is equivalent to
$$\lim_{h \to 0} \frac{f(a+h)-f(a)-f'(a) h}{|h|}  = 0
$$
Good so far.
Now let's compare this with the new definition of the multivariable derivative that you are learning, specialized to the case $m=n=1$. To say that $f : \mathbb R \to \mathbb R$ is differentiable at $a$ means that there exists a linear function $x \mapsto Ax$ (where $A$ is a $1 \times 1$ matrix) such that
$$\lim_{h \to 0} \frac{f(a+h)-f(a)- Ah}{|h|} = 0
$$
Well, these two definitions are equivalent, using the $1 \times 1$ matrix
$$A = (f'(a))
$$
To summarize, one reason the multivariable definition of differentiability is written the way it is so as to generalize the 1-variable definition.
As for our second question, one key feature of a differentiable real valued function $f : \mathbb R^n \to \mathbb R$ is that the directional derivatives at $a$, and in particular partial derivatives which are a special case of directional derivatives, may be evaluated by a useful formula expressed in terms of the $m \times 1$ matrix $A$, assuming that $f$ is differentiable: for any unit vector $\vec v$ one has
$$D_v f(a) = A(v)
$$
So for example if $a = \vec e_i$ is the unit vector in the $i$th coordinate direction, on the left hand side one gets the usual $\frac{\partial f}{\partial x_i}(a)$ and on the right hand side one gets the $i$th coordinate of the matrix $A$. Thus the matrix is the same as the gradient vector of $f$.
But if $f$ is not differentiable at $a$ then all those nice relationships between partial derivatives and the matrix $A$ are out the window, because of course $A$ does not exist. You should definitely crack open your old multivariable calculus book and look for an example of this phenomenon, i.e. an example of a function $f : \mathbb R \to \mathbb R$ and a point $a \in \mathbb R$ such that all partial derivatives $\frac{\partial f}{\partial x_i}(a)$ exist, and yet $f$ is not differentiable at $a$.
On the other hand, if all of the first order partial derivatives of $f : \mathbb R^m \to \mathbb R$ exist, and if each of the partial derivative functions $\frac{\partial f}{\partial x_i} : \mathbb R^m \to \mathbb R$ is continuous, then a miracle occurs: $f$ is differentiable! That's one of the basic theorems that you'll find in an advanced multivariable calculus book.
So, to answer your final question:  It is possible for directional derivatives to exist even when the function is not differentiable at a particular vector? Yes, it is possible, but only when at least one of the directional derivative functions is not continuous.
