There is a quadratic form $q$ on a finite dimensional real vector space with associated symmetric bilinear form $\phi$. There is a subspace $R \leq V$ such that $\phi(r,v) = 0 \forall r \in R$ and $\forall v \in V$. I want to show that $q'(r+R) = q(r)$ defines a quadratic form on the quotient space $V/R$.
The reason why I'm struggling on this question is because I don't understand what this actually amounts to showing. What do I have to do? Can't we just restrict $\phi$ to $V/R$ and then the restriction of $\phi|_{V/R}$ to its diagonal is the quadratic form $q'$ on the quotient space? What do we actually have to do/show?
You can't "restrict $\phi$ to $V/R$" because $V/R$ is not a subset of $V$.
What you need to show is that $q(v) = q(v')$ for all $v, v' \in V$ such that $v + R = v' + R$; this is what is required for $q'$ to be well-defined. Once we know $q'$ actually exists then we check that it is a quadratic form, which is straightforward.
Put another way, given the canonical projection $\pi : V \to V/R$ you need to show that there exists $q' : V/R \to \mathbb R$ such that $q'\circ\pi = q$; this requires the specific properties of $R$. By linearity and surjectivity of $\pi$ it then follows that $q'$ is a quadratic form.