Confusion over this definition of a tensor I see this definition sometimes for tensors, this specific wording comes from "The Poor Man's Introduction to Tensors" by Justin C. Feng:

Simply put, a tensor is a mathematical construction that “eats” a
  bunch of vectors, and “spits out” a scalar. The central principle of
  tensor analysis lies in the simple, almost trivial fact that scalars
  are unaffected by coordinate transformations. From this trivial fact,
  one may obtain the main result of tensor analysis: an equation written
  in tensor form is valid in any coordinate system.

I understand that the appeal of tensors is that they don't have a preference for any particular coordinate system and transform predictably when moving between systems. What I don't see is why giving a scalar is part of the definition above, can I not multiply two tensors together and get something that isn't a scalar? For instance (from Wikipedia):
$$A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}A_{\gamma\delta}$$
Does this statement maybe allude to the fact that tensors can be defined as members of vector/dual spaces combined with the tensor product?
 A: A tensor is defined as an object that linearly maps an ordered pair of vectors from the Cartesian product $V\times V$ to scalars, where both vectors in the ordered pair belong to $V$. Hence
\begin{equation}\tag{1}
\bf{T}: \it{V\times V \rightarrow} \mathbb{R}
\end{equation}
Let $v$ and $w$ be vectors that belong to $V$ and $(v,w)$ belong to $V\times V$. Then
$$\bf{T} \it{(v,w)}=a$$
Now if we represent $\bf{T}$ as the tensor product of dual space vectors, $v^*\otimes w^*$ then the above statement is written as (this equation is the definition of tenosr product)
$$\langle v^*,v\rangle \langle w^*,w\rangle =a$$
Hence $a$ is simply $v^*_\mu v^\mu w^*_\nu w^\nu$. The representation of tensor as the tensor product of dual space vectors helps us express this map in terms known objects. Hence a tensor is expressed as
$$\bf{T}=\it{v^*_\mu w^*_\nu e^\mu \otimes e^\nu}$$
Equation $(1)$ is what the author meant when he said that tensor eats a bunch of vectors and spits out a scalar. 
A: I agree, the wording "a tensor is a mathematical construction" etc. is misleading. 
It's my opinion, but I think the book means that one can always (cleverly) contract a tensor so that the result is a scalar. The most obvious contraction is with the basis vectors, and you need as many as the rank of the tensor, so
$$A^{\mu\nu} e_\mu e_\nu$$
$$B^{\mu}_{\rho\sigma} e_\mu e^\rho e^\sigma$$
which are both scalars. Also, I think the book means "an equation written in this particular tensor form is valid in any coordinate system": in fact, a vectorial tensor equation will not be valid in any coordinate system as it is (indeed, it transforms as a vector).
$$T^{\mu\nu} e_\mu e_\nu\rightarrow T^{\alpha\beta} \Lambda^\mu_\alpha \Lambda^\nu_\beta e_\alpha (\Lambda^{-1})^\alpha_\mu e_\beta (\Lambda^{-1})^\beta_\nu=T^{\mu\nu} e_\mu e_\nu$$
$$T^{\mu\nu} e_\mu \rightarrow T^{\alpha\beta} \Lambda^\mu_\alpha \Lambda^\nu_\beta e_\alpha (\Lambda^{-1})^\alpha_\mu = T^{\alpha\beta} \Lambda^\nu_\beta e_\alpha \not=T^{\mu\nu} e_\mu$$
A: 
Basically yes, I don't see how the definition that they give scalars is compatible with the equation above that seems to take two (2,0)-tensors and gives another (2,0)-tensor, not a scalar.

The equation
$$A^{\alpha \beta} = g^{\alpha \gamma} g^{\beta \delta} A_{\gamma \delta}$$
does not contain any tensors.  All of the terms in that equation are the components of tensors in a particular basis, and the distinction is crucial.
There is a lot to unpack here - you may find that my preliminaries in my answer here may be enlightening.  I also go into considerable detail in the first part of this answer. 
In any case, let's consider the equation $$A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta} B_{\gamma\delta}$$
This corresponds a tensor $\mathbf A$ which eats two covectors and spits out a real number.  It's action can be summarized as follows:


*

*The first covector $\boldsymbol \omega$ is mapped to its vector counterpart $\boldsymbol{\omega}^\sharp$ through the use of the inverse metric, i.e. $\boldsymbol \omega^\sharp = \mathbf{g^{-1}}(\boldsymbol \omega, \bullet)$

*The second covector $\boldsymbol \sigma$ is similarly mapped to its vector counterpart $\boldsymbol \sigma^\sharp = \mathbf{g^{-1}}(\boldsymbol \sigma,\bullet)$

*The resulting vectors are both fed to the $(0,2)$-tensor $\mathbf B$
so we have
$$\mathbf A(\boldsymbol \omega,\boldsymbol \sigma) := \mathbf B(\boldsymbol \omega^\sharp,\boldsymbol \sigma^\sharp)$$
Expressing this in component form, we have that
$$(\omega^\sharp)^\gamma = g^{\alpha \gamma} \omega_\alpha$$
$$(\sigma^\sharp)^\delta = g^{\beta \delta} \sigma_\beta$$
and so
$$\mathbf A(\boldsymbol \omega,\boldsymbol \sigma) = \mathbf B\left((\omega^\sharp)^\gamma \hat e_\gamma, (\sigma^\sharp)^\delta \hat e_\delta\right)$$
$$ = (\omega^\sharp)^\gamma (\sigma^\sharp)^\delta \mathbf B(\hat e_\gamma,\hat e_\delta) = g^{\alpha \gamma} \omega_\alpha g^{\beta \delta} \sigma_\beta B_{\gamma \delta} = A^{\alpha \beta} \omega_\alpha \sigma_\beta$$
Therefore we can calculate the components of $\mathbf A$ to be
$$A^{\alpha \beta} = g^{\alpha\gamma}g^{\beta\delta} B_{\gamma\delta}$$

I'm sure you noticed that I used a $B$ instead of an $A$ in the last term of your equation - I did this to emphasize that the object on the left (with upstairs indices) is not the same as the object on the right (which has downstairs indices).  They are related to each other, but are different.  It is common convention to make notation simpler by using the letter $A$ for both objects and to simply distinguish them by where you place the indices, but this can lead to enormous confusion for beginning students.
A: From the comments:

I don't see how the definition that they give scalars is compatible with the equation above that seems to take two (2,0)-tensors and gives another (2,0)-tensor, not a scalar. 

It's because you're also combining things with another tensor operation, namely contraction.    Specifically, the contraction of a $(1,1)$ tensor $T: V \times V^* \to \mathbb{R}$ is defined as
$$
CT = \sum_\sigma T({v^\sigma}^*, v_\sigma)
$$
where $\{v_\sigma\}$ is a basis for $V$, and $\{{v^\sigma}^*\}$ is the corresponding dual basis for $V^*$.  It is a standard exercise to show that this definition is independent of the basis chosen;  the proof hinges on the definition of the dual basis and the linearity of any tensor with respect to its arguments.  In abstract index notation, the contraction of the tensor $T^a {}_b$ is written $T^a {}_a$.  This notion can be extended to define the contraction of a $(n,m)$ tensor;  it yields an $(n-1,m-1)$ tensor as we would expect.
So when we write down the product
$$
A^{ab} = g^{ac} g^{bd} A_{cd},
$$
we are really implicitly doing the following:


*

*Define the $(4,2)$ tensor $g^{ae} g^{bf} A_{cd}$, whose value on a set of four dual vectors and two vectors is obtained by feeding the first two dual-vector arguments into one "copy" of $g$, the second two dual-vector arguments into the second "copy" of $g$, and the two vector arguments into $A$, and then multiplying everything together.

*Perform the contraction between the second dual vector "slot" and the first vector "slot", as well as the contraction between the fourth dual vector "slot" and the second vector "slot".  This yields a tensor of type (2,0) as expected.


The result can still be viewed as a map from $V^* \times V^* \to \mathbb{R}$:  we would simply "feed in" two more dual vectors into the remaining two uncontracted "slots" in the original expression $g^{ac} g^{bd} A_{cd}$.
A: A tensor transforms like a scalar. When you transform the tensor basis to new coordinate system, the tensor coefficients are transformed by the inverse transformation, and the tensor remains invariant.
For instance, without loss of generality, let  
$$t= t^{i}_{\;j} \frac{\partial}{\partial u_{i}} \otimes du^{j}$$
$$\frac{\partial}{\partial u_{i}}=\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$
$$du^{j}=\frac{\partial u_{j}}{\partial w_{l}} dw^{l}$$
$$t= t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}\otimes\frac{\partial u_{j}}{\partial w_{l}} dw^{l}$$
$$t= t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial u_{j}}{\partial w_{l}}\frac{\partial}{\partial w_{k}}\otimes dw^{l}$$
$$t= t^{k}_{\;l} \frac{\partial}{\partial w_{k}}\otimes dw^{l}$$
where
$$t^{k}_{\;l} =t^{i}_{\;j} \frac{\partial w_{k}}{\partial u_{i}} \frac{\partial u_{j}}{\partial w_{l}}$$
$t$ transforms like scalar and it's valid in any coordinate system.
EDIT: added the $(1,0)$ transformation based on Mauro Giliberti comment.
Let $$v=v^{i}\frac{\partial}{\partial u_{i}}$$ then $$\frac{\partial}{\partial u_{i}}=\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$
$$v=v^{i}\frac{\partial w_{k}}{\partial u_{i}} \frac{\partial}{\partial w_{k}}$$
$$v=v^{k} \frac{\partial}{\partial w_{k}}$$
which transforms like a tensor.
