(I'm going to assume that we're talking about real finite-dimensional vector spaces here, although what I'm about to say mostly generalizes.)
Every vector space $V$ has a dual $V^*=Hom(V,\mathbb{R})=\{\mbox{ linear }x:V\to\mathbb{R}\}$.
Exercise: $V^*$ is a vector space.
In fact, $V^*$ is a finite-dimensional vector space. See this by choosing a basis $\{b_i\}$ of $V$, and defining $\beta_i$ to be the linear function such that $\beta_i(b_j)=\delta_{ij}$.
Exercise: The $\beta_i$ form a basis of $V^*$.
These form the "dual basis" of $\{b_i\}$, and of course implies that $V^*$ has the same dimension as $V$.
Exercise: A choice of basis induces an isomorphism $V\leftrightarrow V^*$.
But a priori any isomorphism depends on a choice of basis! $V$ is isomorphic to $V^*$, but without extra structure, one cannot say that $V$ is equal to $V^*$.
A bilinear form $\langle\cdot,\cdot\rangle$ is a bilinear map* $V\times V\to\mathbb{R}$. The dot product on $\mathbb{R}^n$ is a bilinear form. A choice of symmetric, nondegenerate* bilinear form determines an isomorphism $V\leftrightarrow V^*$ via $v\mapsto \langle -,v\rangle$.
*N.B.: Your professor's definition of "bilinear form" may vary from mine here. The term is often taken to mean symmetric and positive-definite right out of the box.
Exercise: The map $v\mapsto\langle-,v\rangle$ induced by a symmetric, nondegenerate bilinear form is an isomorphism.
You'll want to make your peace someday with tensor algebras, which are the natural homes of symmetric bilinear forms. Introducing you to them is beyond the scope of this post.
For an abstract, perhaps unreadable, introduction, you might consult chapter 2 of Warner, Introduction to Differentiable Manifolds and Lie Groups. I would myself welcome any suggestions of advanced linear algebra textbooks.