$X_1, X_2, \ldots, X_n$ are iid $N(0,\theta), 0 < \theta < \infty$
Show $$\sum_{i=1}^{n} X_i^2$$
is a sufficient statistic for $\theta$.
My attempt at this is
$S = (X_1^2 + X_2^2+\cdots+X_n^2)/n$
$$E[S]= \frac{1}{n} E[X_1^2+X_2^2+\cdots+X_n^2]$$
$$E[X^2 - [E(X)]^2 = \operatorname{var}X = \theta$$
$$E[X^2] = \theta = \frac{1}{n}(nt) = t$$
So $S$ is an unbiased estimator of $t$.
$$\operatorname{var}(S) = \frac 1 {n^2} [\operatorname{var} X_1^2 + \operatorname{var} X_2^2 +\cdots+ \operatorname{var}X_n^2]$$
$$\operatorname{Var}(X^2) = E(X^4) - (E[X^2])^2$$
Since $E(X^2) = t$, we must find $E(X^4)$
Using moment generating function for $X$, $e^{\frac{ts^2}{2}}$
Exanding, $s^4 = 1 + (1/2)ts^2 + (1/8)t^2s^4$
coeff of $s^4$ is $t^2/8 = E(X^4/4!)$,
therefore $E(X^4) = 3t^2$
$$\operatorname{var}(X^2)=\frac{3t^2}{2-t^2} = \frac{t^2}2 $$
and $\operatorname{var}(S) = \dfrac{t^2}{2n}$
Is this a sufficient/correct explanation?
Everything you write seems irrelevant to the question.
To say that $X_1^2+\cdots+X_n^2$ is a sufficient statistic for $\theta$ means that the conditional distribution of $(X_1,\ldots,X_n)$ given $X_1^2+\cdots+X_n^2$ does not depend on $\theta$.
That is what you need to show.
You might want to use Fisher's factorization theorem: $t(X_1,\ldots,X_n)$ is sufficient for $\theta$ if the density can be factored as $$ f(x_1,\ldots,x_n \mid \theta) = g(t(x_1,\ldots,x_n)\mid\theta) h(x_1,\ldots,x_n). $$ where the second factor does not depend on $\theta$.
(In the case of the Poisson distribution the second factor is non-trivial, but for the family of distributions you have here I think you'll get $1$.)