The following relies only on the fact that $X_1,\ldots,X_n$ are i.i.d with finite mean $\theta$, finite variance $\sigma^2(>0)$ and $E|\delta_n|<\infty$ for every $n\ge 1$.
Your $\delta_n$ is a U-statistic, since it can be written in the form
$$\delta_n=\frac1{\binom{n}{2}}\sum_{1\le \alpha<\beta\le n}X_\alpha X_\beta$$
Asymptotic distributions of U-statistics in general can be derived using projections.
If $\theta\ne 0$, the general result gives
$$\sqrt n(\delta_n-\theta^2)\stackrel{d}\longrightarrow N(0,4\theta^2\sigma^2) \tag{1}$$
And if $\theta=0$,
$$n(\delta_n-\theta^2)\stackrel{d}\longrightarrow \sigma^2(\chi^2_1-1) \tag{2}$$
To prove $(1)$ using projection theorem, let $T_n=\sqrt n(\delta_n-\theta^2)$, so that $E(T_n)=0$.
Define $K_P(X_i)=E(T_n\mid X_i)$, which simplifies to
$$K_P(X_i)=\frac{\sqrt n}{\binom{n}{2}}(n-1)(\phi_1(X_i)-\theta^2)=\frac2{\sqrt n}(\phi_1(X_i)-\theta^2)\,,$$
with $\phi_1(X_1)=E[X_1X_2\mid X_1]=\theta X_1$.
Note that $E(\phi_1)=\theta^2$ and $\operatorname{Var}(\phi_1)=\theta^2\sigma^2=\sigma_1^2$ (say).
So by classical CLT, $$V_P=\sum_{i=1}^n K_P(X_i) \stackrel{d}\longrightarrow N(0,4\sigma_1^2)$$
It can be shown that $\operatorname{Var}(\delta_n)=\frac{4\sigma_1^2}{n}+O\left(\frac1{n^2}\right)$. And clearly, $\operatorname{Var}(V_P)=4\sigma_1^2$.
Therefore, as $n\to \infty$, $$\operatorname{Var}(T_n)-\operatorname{Var}(V_P)=O\left(\frac1n\right)\longrightarrow 0$$
Projection theorem (which is another application of Slutsky's theorem) then says that limiting distributions of $T_n$ and $V_P$ are identical.
Edit:
I might have complicated things before. Ignore the proof above as it is missing some details.
For $\theta \ne 0$, simply write
$$\sqrt n(\delta_n-\theta^2)=\sqrt n\left(\overline X^2 -\theta^2\right)-\frac1{\sqrt n}\left(\frac{S^2}{n-1}\right)$$
Using delta method, OP can show that
$$\sqrt n\left(\overline X^2 -\theta^2\right)\stackrel{d}\longrightarrow N(0,4\theta^2\sigma^2)$$
And we know that
$$\frac{S^2}{n-1} \stackrel{P}\longrightarrow \sigma^2\,,$$
which implies $$\frac1{\sqrt n}\left(\frac{S^2}{n-1}\right)\stackrel{P}\longrightarrow 0$$
Hence $(1)$ follows directly from Slutsky's theorem.
Now when $\theta=0$,
$$n(\delta_n-\theta^2)=n\overline X^2 - \frac{S^2}{n-1}$$
We have from CLT and continuous mapping theorem that
$$n\overline X^2\stackrel{d}\longrightarrow \sigma^2\chi^2_1$$
And as before, $$\frac{S^2}{n-1} \stackrel{P}\longrightarrow \sigma^2$$
Therefore, Slutsky's theorem again proves $(2)$.
