Let $R$ be a unital commutative ring, and $M$ an $R$-module. One of the most useful ways of defining a quadratic form is as the diagonal of a bilinear form:
Definition 1: A map $Q \colon M \to R$ is called a quadratic form if there exists some $R$-bilinear form $B \colon M \times M \to R$ such that $Q(m) = B(m, m)$ for all $m \in M$.
There is another more axiomatic definition, explicitly requiring that $Q$ behaves like a homogeneous polynomial of degree 2.
Definition 2: A map $Q \colon M \to R$ is called a quadratic form if $Q(rm) = r^2 m$ for all $r \in R$ and $m \in M$, and the polar form $Q_P \colon M \times M \to R$ is a bilinear form, where $Q_P(m, n) = Q(m + n) – Q(m) – Q(n)$.
It is very easy to show that quadratic form in the sense of Definition 1 is a quadratic form in the sense of Definition 2, but I'm finding it difficult to show the converse.
Question: Is there a simple proof that Definition 2 implies Definition 1 in general?
I can prove this with some further assumptions. The first easy case is if $2$ is invertible in $R$, since then we have that $\frac{1}{2} Q_P$ is a bilinear form such that $\frac{1}{2} Q_P(m, m) = Q(m)$.
The second easy case is if $M$ is free on some basis $\{e_i \mid i \in I\}$ where $I$ is totally ordered. (As far as I can tell, this covers basically every case that people actually use "in the real world"). We then define a bilinear form $B \colon M \times M \to R$ by setting
$$ B(e_i, e_j) = \begin{cases}
Q_P(e_i, e_j) & \text{if } i < j, \\
Q(e_i) & \text{if } i = j, \\
0 & \text{if } i > j.
\end{cases}$$
Together with the fact that
$$ Q(m_1 + \cdots + m_n) = \sum_i Q(m_i) + \sum_{i < j} Q_P(m_i, m_j), $$
it is easily seen that $B$ is indeed a bilinear form on $M$ such that $B(m, m) = Q(m)$ for all $m \in M$.
Best Answer
They are not equivalent. We'll say that $Q$ admits a bilinear refinement if there exists bilinear $B$ such that $Q(m) = B(m, m)$. Let $R = \mathbb{Z}/4$ and $M = \mathbb{Z}/2$, and consider the function $Q : M \to R$ given by
$$Q(0) = 0, Q(1) = 1.$$
We have $Q(rm) = r^2 Q(m)$ (this only imposes the condition that $Q(0) = 0$). The polar form, which I'll write $B_Q$, satisfies
$$B_Q(0, 0) = 0$$ $$B_Q(1, 0) = 0$$ $$B_Q(0, 1) = 0$$ $$B_Q(1, 1) = 2.$$
We can check that a map $B : M \times M \to R$ is bilinear iff $B(0, 0) = B(1, 0) = B(0, 1) = 0$ and $B(1, 1) \in 2 \mathbb{Z}/4$ (because the only nontrivial condition linearity imposes here is that $B(-1, 1) = B(1, -1) = -B(1, 1)$), which is the case here. So $Q$ is a quadratic form.
But $Q$ does not admit a bilinear refinement: any bilinear form $B$ must have image in $2 \mathbb{Z}/4$ and hence so must its diagonal quadratic form, which $Q$ doesn't.
The difference between these two definitions actually matters "in the real world," and the second definition turns out to be the correct one. For example, there is a cohomology operation called Pontryagin square $H^{2k}(-, \mathbb{Z}/2) \to H^{4k}(-, \mathbb{Z}/4)$, which is quadratic. It's a lift of the cup square, to which it reduces $\bmod 2$, and I believe it does not admit a bilinear refinement, although I don't know enough about how to compute it to be sure.
For another example, given two abelian groups $\pi_2, \pi_3$ we can define a quadratic map $\pi_2 \to \pi_3$, which is a slight generalization of your second definition. If $\pi_2, \pi_3$ are the second and third homotopy groups of a space then there is a natural homotopy operation $\pi_2 \to \pi_3$ given by precomposing with the Hopf fibration. This map is a quadratic refinement of the Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$, and I believe it also does not admit a bilinear refinement. If $X$ is a space whose only nonvanishing homotopy groups are $\pi_2$ and $\pi_3$ then it is classified by a $k$-invariant $k \in H^4(B^2 \pi_2, \pi_3)$, and it turns out that this cohomology group can naturally be identified with the group of quadratic maps $\pi_2 \to \pi_3$ (this is due to Eilenberg and Mac Lane). In other words this homotopy operation completely classifies spaces whose only nonvanishing homotopy groups are $\pi_2, \pi_3$. Equivalently, it classifies braided monoidal grouplike groupoids (this is a mouthful but it's really a very natural thing to consider).
When $k = 1$ the Pontryagin square $H^2(-, \mathbb{Z}/2) \to H^4(-, \mathbb{Z}/4)$ is classified by a map $B^2 \mathbb{Z}/2 \to B^4 \mathbb{Z}/4$ and hence by a cohomology class in $H^4(B^2 \mathbb{Z}/2, \mathbb{Z}/4)$ which I would be willing to bet corresponds to the quadratic form $Q$ above, although I don't know how to prove this.
For more of this sort of thing see, for example, Hopkins and Singer's Quadratic functions in geometry, topology, and M-theory.