Is a bilinear form a tensor or a scalar "output'' of that tensor? As I understand it, a bilinear form, $b$, on a (real) vector space, $V$, is a bilinear function $\;b:V\times V \to \mathbb{R}$.  Let's take the vector space to simply be flat euclidean 3-space, $\mathbb{E}^3$ (and forget about things like covariant and contravariant indices and dual vectors).  The contraction of some 2-tensor $\mathbf{B}$, with two vectors, $\mathbf{u}$ and $\mathbf{v}$, seems to meet the definition of a bilinear form:
$$
\text{for }\;\; \mathbf{u}=u_i\hat{\mathbf{e}}_i \;,\;\mathbf{v}=v_i\hat{\mathbf{e}}_i\,\in \mathbb{E}^3 \;\;,\;\; \mathbf{B}=B_{ij}\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j\,\in \mathbb{E}^3\otimes\mathbb{E}^3 \qquad \text{then} \qquad
\qquad\mathbf{u}\cdot\mathbf{B}\cdot\mathbf{v} = B_{ij}u_iv_j \in \mathbb{R}
$$
(einstein summation) where $\hat{\mathbf{e}}_i$ (for i=1,2,3) is some arbitrary orthonormal basis for $\mathbb{E}^3$. (I think some would write the right-hand-side of the above as $\mathbf{B}(\mathbf{u},\mathbf{v})$)
My question is this: in the above example, this is the 2-tensor $\mathbf{B}$ the bilinear form? or is the full expression $\mathbf{u}\cdot\mathbf{B}\cdot\mathbf{v}$ the bilinear form?  If a bilinear form on $\mathbb{E}^3$ defined as a bilinear function $\;b:\mathbb{E}^3\times \mathbb{E}^3 \to \mathbb{R}$, doesn't this notion mean that $b$ itself must be an element of $\mathbb{R}$? or is that wrong?  The 2-tensor $\mathbf{B}\notin \mathbb{R}$. But $\mathbf{u}\cdot\mathbf{B}\cdot\mathbf{v}\in \mathbb{R}$. Which is the bilinear form?
Note: my BS is in physics and I'm currently an engineering PhD student. I have worked with (cartesian) tensors and linear algebra throughout, but it was never done in a formal way using "proper" mathematical terms or notation. I never even heard the term "bilinear form" until very recently even though I've undoubtedly encountered them thousands of times.
 A: My first remark for you is to NOT overlook the distinction between a vector space and its dual: if you do, then you're going to make things a lot harder on yourself in the medium-long run.

A bilinear form on $V$ is as you said in your first sentence, a function $b:V\times V\to\Bbb{R}$. I don't know where you're getting the rest of the stuff from. You say

doesn't this notion mean that $b$ itself must be an element of $\Bbb{R}$?

Absolutely not! It is a function $V\times V\to\Bbb{R}$. Functions are not real numbers. The target space of the function is a real number in this case, but the function itself is obviously not a real number. I'm not sure why you introduced $\mathbf{B}$ in your post. The object $b$ is what is of interest; you can call $b$ a bilinear functional on $V$ or a bilinear form on $V$, or you can also call it a $(0,2)$-tensor on $V$, so $b\in T^0_2(V)$ (again, $b\notin \Bbb{R}$). The words form or functional merely emphasize that the target space is $\Bbb{R}$. The most agnostic way of saying it is that $b$ is a bilinear mapping from $V\times V$ into $\Bbb{R}$; such a phrasing immediately generalizes to situations where you have three different vector spaces $V_1\times V_2\to V_3$.
Next, given two vectors $u,v\in V$, we get a number $b(u,v)\in\Bbb{R}$. We call this number the value/output of the bilinear form/functional/$(0,2)$-tensor $b$ on the pair of vectors $(u,v)$.
Next, you seem to be writing a basis-expansion of $b$, but unfortunately things are not in the right space. As you've currently written it, $b\in V\otimes V$, but as I mentioned in my first sentence, you shouldn't disregard the dual. We have the canonical isomorphism (for finite-dimensional $V$):
\begin{align}
\text{Hom}^2(V\times V;\Bbb{R})&\cong \text{Hom}(V\otimes V;\Bbb{R})=(V\otimes V)^*\cong V^*\otimes V^*.
\end{align}
So, the object $b$ naturally wants to live in $V^*\otimes V^*$, NOT $V\otimes V$.

If you want to write things out using a basis, then fix a basis $\{e_1,\dots,e_n\}$ of $V$ (you don't even need orthonormality), and let $\{\epsilon_1,\dots,\epsilon_n\}$ be the dual basis. Then, you can write
\begin{align}
b&=\sum_{i,j=1}^nb(e_i,e_j)\epsilon_i\otimes\epsilon_j\equiv\sum_{i,j=1}^nb_{ij}\,\epsilon_i\otimes\epsilon_j
\end{align}
So, if $u,v\in V$, then you can write them as $u=\sum\limits_{i=1}^nu_ie_i$ and $v=\sum\limits_{i=1}^n v_ie_i$ for some numbers $u_i,v_i$ (in fact, $u_i=\epsilon_i(u)$, $v_i=\epsilon_i(v)$, i.e the component $u_i\in\Bbb{R}$ equals the value of the covector $\epsilon_i\in V^*$ when evaluated on the vector $u\in V$). Then, the value $b(u,v)$ can be written as:
\begin{align}
b(u,v)&=\sum_{i,j=1}^nb_{ij}\,(\epsilon_i\otimes\epsilon_j)(u,v)\\
&:=\sum_{i,j=1}^nb_{ij}\,\epsilon_i(u)\cdot \epsilon_j(v)\\
&=\sum_{i,j=1}^nb_{ij}u_iv_j.
\end{align}

The relation with matrices is that you can store these numbers $b_{ij},u_i,v_j$ as
\begin{align}
[b]&=\begin{pmatrix}
b_{11} & \cdots & b_{1n}\\
\vdots & \ddots & \vdots\\
b_{n1}&\cdots & b_{nn}
\end{pmatrix},
[u]=
\begin{pmatrix}
u_1\\
\vdots\\
u_n
\end{pmatrix},
\quad\text{and}\quad
[v]=
\begin{pmatrix}
v_1\\
\vdots\\
v_n
\end{pmatrix}
\end{align}
We call $[b]\in M_{n\times n}(\Bbb{R})$ the matrix representation of $b$ relative to the basis $\{e_1,\dots, e_n\}$ of $V$, and we call $[u]$ the coordinate vector of $u$ relative to the basis $\{e_1,\dots, e_n\}$ of $V$ (and likewise for $v$).
Then, $b(u,v)=[u]^t\cdot [b]\cdot [v]$. So, the value of the tensor on a pair of vectors can be carried out using matrix multiplication once you fix a basis $\{e_1,\dots, e_n\}$ of $V$ (but conceptually, bases should be the last thing on your mind).

(I wrote everything above with lower indices simply because you don't like upper indices, so I included explicitly the summation symbols as well; the natural way to write this would have been $\{e_1,\dots,e_n\}$ for the basis, $\{\epsilon^1,\dots,\epsilon^n\}$ for the dual basis, $u=u^ie_i$ and $v=v^je_j$ for the vectors, and $b=b_{ij}\epsilon^i\otimes\epsilon^j$ for the tensor $b$. Note also the dual basis only comes up if you want to abstractly write $b$ in terms of a basis as $b=b(e_i,e_j)\,\epsilon^i\otimes \epsilon^j$).
