Find a basis with respect to which this bilinear form is diagonal

We have the following symmetric bilinear form on $\\Q^4$ (vector space over rational numbers) with respect to standard basis $\{e_1,e_2,e_3,e_4\}$ $$g(v,w)=v^tAw$$ $$A=\begin{bmatrix}1 & 2 &3&4\\2 & 3 & 4 & 5\\3&4&5&6\\4&5&6&7\end{bmatrix}$$

How to find a basis on which g is diagonal. I suppose using Gram-Schmidt algorithm we find orthogonal basis, but could someone explain this explicitly, please?

There is no need to look for an orthogonal basis. Note that our goal is not to diagonalize $A$ using similarity transform, but to diagonalize $A$ using congruence. That is, what we are looking for is not an invertible matrix $P$ such that $P^{-1}AP$ is diagonal, but an invertible $P$ such that $P^TAP$ is diagonal.

(When $P$ is a real orthogonal matrix, $P^{-1}AP$ and $P^TAP$ coincide, but it is not always possible to find such an orthogonal $P$ over the rationals and this is not our goal anyway.)

You may apply simultaneous elementary row and column reductions to obtain $P$. More specificically, let $$ E_1=\begin{pmatrix}1&0&0&0\\-2&1&0&0\\-3&0&1&0\\-4&0&0&1\end{pmatrix}. $$ Then $$ B=E_1AE_1^T=\begin{pmatrix}1&0&0&0\\0&-1&-2&-3\\0&-2&-4&-6\\0&-3&-6&-9\end{pmatrix}. $$ Now let $$ E_2=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&-2&1&0\\0&-3&0&1\end{pmatrix}. $$ Then $$ D=E_2BE_2^T=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}. $$ Now $D=(E_2E_1)A(E_1^TE_2^T)$. So you may take the basis vectors as the column vectors of $P=E_1^TE_2^T$.