Let $E$ be an elliptic curve over a finite field $\mathbb{F}_q$ where $q$ is prime. Let $P$ be a point on $E$.
Consider the point $Q=(q+1)P=P+\cdots+P$, which is $P$ added to itself $q+1$ times. Due to the fact that we are in the finite field $\mathbb{F}_q$, I would expect the "scalar", in this case, $q+1$, to respect the ground field property and have $Q=P$, but this is not the case.
I am told that $E(\mathbb{F}_q)$ is either cyclic or a product of two cyclic groups, and that the order may not be $q$. Are there underlying concepts that I may be missing?
You are right; you are missing some points. As of now you seem to be confusing the torsion points of elliptic curves with the (torsion elements of) multiplicative group $\mathbb F_q^*$. This is not a bad analogy at a higher level; but it is not superficially true. You will need to spend quite some time to understand everything. You can find a rigorous proof of this fact in Silverman's book on elliptic curves. If you want to understand heuristically, here is the following.
An elliptic curve over $\mathbb C$ is isomorphic to $S^1 \oplus S^1$ as a topological group, where $S^1$ is the circle group. The $m$-torsion has size $\mathbb Z/m\mathbb Z\oplus\mathbb Z/m\mathbb Z$. Due to the fact that elliptic curves are actually algebraic, and in particular, the torsion points are algebraic, these facts largely carry over to elliptic curves over algebraically closed char-$p$ fields(Here is where I am fudging things). The $\mathbb F_q$-rational points are obviously finite in number and they belong to some $m$-torsion group for the elliptic curve considered over the algebraic closure of $\mathbb F_q$. That it is either cyclic or a direct sum of two cyclic group now follows from the structure theorem of finitely generated abelian groups.