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.
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
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 $b=b(e_i,e_j)\,\epsilon^i\otimes \epsilon^j$).