I was reading some notes on elliptic curves and abelian varieties and for an elliptic curve $E$ over a field $k$, I saw the following result: $$\operatorname{Isog}(E)\longleftrightarrow \{K\subset E \mid K \text{ is a finite subgroup}\}$$ (here $\operatorname{Isog}(E)$ is the set of all isogenies from $E$ upto isomorphism and the two sided arrow implies bijection)
I have seen the result that any if we have an isogeny $\phi: E\rightarrow E'$, then there is a finite subgroup $K\subset E$ such that such that $E'\cong E/K$ and $\phi$ represents the quotient map. So, I understand why this result should be true. But, I do not understand what finite subgroup means in this case. Is it saying that $K$ is finite over $k$? That is my guess because I have seen that over an algebrically closed field and for $char(k)\not\mid n$, we have $E[n]$ is just $(\mathbb{Z}/n\mathbb{Z})^2$. This might seem like a silly question, but my understanding of group scheme is not great. Is my interpretation correct?
Your interpretation is correct.
Given a base scheme $S$ by a finite group scheme what is almost invariably meant is a group scheme $G\to S$ which finite and locally free. If $S$ is locally Noetherian then this is equivalent to the condition that $G\to S$ is a group scheme which is finite and flat. Of course, if $S=\mathrm{Spec}(k)$ then this all collapses to just saying that $G\to\mathrm{Spec}(k)$ is an affine group scheme such that $\mathcal{O}(G)$ is a finite-dimensional $k$-space.
So then, by $K\subseteq E$ a finite subgroup scheme, what is meant is an (equivalence class of) homomorphisms $K\to E$ which are closed embeddings (equiv. have trivial kernel), and for which $K$ is a finite group scheme (over $\mathrm{Spec}(k)$). In particular, as you noted one can also characterize such $K$ as precisely the (closed) subgroup schemes of $E[n]$ for some $n\geqslant 1$.