This is a great question.

Have you seen Manin’s book “Cubic forms: algebra, geometry, arithmetic”?

Here are a two basic comparisons between quadratic forms and higher-degree forms.

1. Diagonalizability.

Outside characteristic 2, a quadratic form can be diagonalized after a linear change of variables, but for $n \geq 3$, a form of degree $n$ might not be diagonalizable after any linear change of variables.
While any binary cubic form over $\mathbf C$ can be diagonalized, in more than two variables cubic forms over $\mathbf C$ need not be diagonalizable. For example, 
the three-variable cubic form $x^3 - y^2z - xz^2$ can't be diagonalized over $\mathbf C$ for a reason related to elliptic curves: see my comments on the MO pages [here][1] and [here][2].
For each $d \geq 3$ and $n \geq 2$ *except* for $d=3$ and $n = 2$, there are diagonalizable forms of degree $d$ in $n$ variables over $\mathbf C$. 

2. Group theory.

For a form $f(x_1,\ldots,x_n)$ over a field $F$, its *orthogonal group* is the linear changes of variables on $F^n$ that preserve it:
$$
O(f) = \{A \in {\rm GL}_n(F) : f(A\mathbf v) = f(\mathbf v) \ {\rm for \ all } \ \mathbf v \in F^n\}.
$$
Quadratic forms have a rich orthogonal group (many reflections).  But 
for a "nondegenerate" form $f$ of degree $d \geq 3$ over an algebraically closed field of characteristic $0$, $O(f)$ is a *finite* group. The version of this over $\mathbf C$ goes back to Jordan.
I believe it also holds in characteristic $p$ when $p > d$.
When $f = x_1^d + \cdots + x_n^d$ for $d \geq 3$, its orthogonal group over $\mathbf C$ is a finite group: the compositions of $n!$ coordinate permutations and scaling of the $n$ coordinates by $d$th roots of unity: a finite group of order $d^n n!$.  That's a big difference between $x^2 + y^2$ and $x^d + y^d$ for $d \geq 3$.








[1]:https://mathoverflow.net/questions/238901/cubic-forms-and-finiteness-of-k-k3

[2]:https://mathoverflow.net/questions/77702/transformation-of-a-cubic-form