Is there something known about the values of the Kronecker symbol $\left(\frac{a_n}{a_{n+1}}\right)$ if $a_n$ is a recursive sequence? I know something about this if $a_n$ is the Fibonacci sequence or the Lucase sequence and for a given sequence, these symbols can be computed. Nevertheless, it would be great to have some results that hold for all recursive sequences of the form $a_{n+1} = \alpha \cdot a_n + \beta \cdot a_{n-1}$ for some $\alpha, \beta$ independent (or even dependent) of $n$.

At the end of his 2012 paper A. Granville states that finding a "usable" formula is an open problem for (certain) Lucas sequences. Thus there does not seem to be much hope to answer your question with $\alpha,\beta$ depending on $n$ in full generality.

In the cited paper I challenged the reader to find a general usable formula for $(x_m/x_n)$. However this special case should be doable. It seems to me you can simply proceed by induction, though you will need to write $x_n = 2^{k_n} y_n$ where one will need some idea of the value of each $k_n$, and each $y_n \mod 8$. But if one has succeeded with the Fibonacci numbers, these issues arise, so I would guess that that method should generalize with some work.

To give an easy general class of example, I work with $x_{n+2}=ax_{n+1}+bx_n$ for all $n\geq 0$, where we suppose $4$ divides $b$, and $a$ and $x_1$ are odd (else the whole sequence will be even). From the recurrence we have $x_{n+2}\equiv ax_{n+1} \pmod b$, and so $x_n\equiv a^{n-1}x_1 \pmod b$ for all $n\geq 1$. This implies that each $x_n$ is odd, and at least one of $x_n, x_{n+1}$ is $\equiv 1 \pmod 4$ and therefore $(x_n/x_{n+1})=(x_{n+1}/x_n)$ for all $n\geq 1$. Moreover $x_{n+1}\equiv bx_{n-1} \pmod {x_n}$ by the recurrence relation and so $(x_{n+1}/x_n)=(b/x_n)(x_{n-1}/x_n)$. Putting these together yields

$(x_n/x_{n+1})= (b/x_n)(x_{n-1}/x_n)$ for all $n\geq 1$.

Proceeding by induction then yields (assuming $x_0\ne 0$ -- if it is we stop one step earlier)

$(x_n/x_{n+1})= (b/x_nx_{n-1}...x_1)(x_0/x_1)$

As $4|b$ the Kronecker symbol $(b/.)$ has period dividing $b$, and so we use that $x_nx_{n-1}...x_1 \equiv a^{n(n-1)/2}x_1^n \pmod b$. Therefore

$(x_n/x_{n+1})= (b/a)^{n(n-1)/2} \ (b/x_1)^n \ (x_0/x_1)$, as desired.

This formula depends only on the value of $n\pmod 4$ and so we can write

$(x_n/x_{n+1})= (x_r/x_{r+1})$, where $r$ is the least residue of $n \pmod 4$.