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Find a diagonal form of the quadratic form
$$f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j.$$

It turned out to be such a problem:
How to change quadratic form $f(x) = sum_{i = 1}^nx_i^2 + sum_{i < j}x_ix_j$ into diagonal form?

We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).

Is there any other way?

The matrix $M$ of this quadratic form has the form
$$
M=frac12(I_n+J_n),
$$
where $J_n$ is the $ntimes n$ matrix with all ones.

The eigenvalues of $J_n$ are obviously $lambda_1=n$ (multiplicity one) and
$lambda_2=lambda_3=cdots=lambda_n=0$ (multiplicity $n-1$). Therefore the eigenvalues of $M$ are $(n+1)/2$ (multiplicity one) and $1/2$ (multiplicity $n-1$). Implying that there exists an orthogonal coordinate transformation $(x_1,x_2,ldots,x_n)mapsto (x_1',x_2',ldots,x_n')$ such that in the primed coordinates the quadratic form takes the form
$$
Q(x_1',x_2',ldots,x_n')=frac12left((n+1)x_1'^2+x_2'^2+x_3'^2+cdots+x_n'^2right).
$$
An orthonormal basis corresponding to the primed coordinates can be found by orthonormalizing any basis of the $(n-1)$-dimensional subspace
$$V={(x_1,x_2,ldots,x_n)inBbb{R}^nmid x_1+x_2+ldots+x_n=0}$$
which is equal to the eigenspace of $J_n$ belonging to the eigenvalue zero. The missing basis vector is the unit vector $dfrac1{sqrt n}(1,1,ldots,1)$ spanning the orthogonal complement of $V$.

It is (very likely) possible to describe an orthonormal basis of $V$ with a lot of symmetries. I'm not sure I want to go there, though. I would start with a complex basis consisting of roots of unity of order $n$, and go from there.

This is too long for the comment box.

The essence of your exercise is given in the following example.

Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \
&= left( x + dfrac{y}{2} + dfrac{z}{2} right)^2 - frac{y^2}{4} - frac{z^2}{4} -frac{yz}{2} + y^2 + z^2 + yz\
&=u^2 + dfrac{3y^2}{4} + dfrac{yz}{2} + dfrac{3z^2}{4}
end{align*},$$
where $u = x + dfrac{y}{2} + dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$begin{align*}
f(x, y, z) &= u^2 + left( dfrac{sqrt{3}y}{2} + frac{z}{2sqrt{3}} right)^2-dfrac{z^2}{12}+dfrac{3z^2}{4} = u^2 + v^2+dfrac{5}{6}z^2
end{align*},$$
which is a required representation of $f.$

The case with $x_1, ldots, x_p$ is just guessing a pattern and using induction (I am guessing).