Is this quotient space even defined? If I add some $v \in V$ to some $u \in U$, I would be adding a polynomial and a constant for the first coordinate (possible) and an ordered pair and a constant for the second coordinate, which does not seem possible.
My intuition is telling me that if we mod out $\mathbb{R}^2$ we should be left with $\mathcal{P}_2(\mathbb{R})$, which aligns with the fact that the dimension of $V/U$ here is 3 since dim $V=5$ and dim $U=2$, but I'm not confident in that, and more importantly I'm not fully grasping how we arrive at that result either way.
Am I missing something fundamental here? I studied quotient groups in group theory last semester and did well with them, but this linear algebra context is throwing me for a loop for some reason. Appreciate any insight.
The vector space $U= \mathbb{R}^2$ is not a subspace of $V = \mathcal{P}_2 \times \mathbb{R}^2$ for the reasons you say: every element of $V$ is of the form $(p,u)$ for some $p \in \mathcal{P}_2$, $u \in U$. But elements of $U$ are not of that form. Since $U$ is not a subspace of $V$, the quotient space $V/U$ is indeed not well-defined.
But your intuition is just about right, and you can patch it up by considering instead $U' := \{0\} \times \mathbb{R}^2$, which is a subspace of $V$. Then an application of the first isomorphism theorem for vector spaces (more or less the same as you learned for groups) gives $$ \mathcal{P}_2 \cong V / U'$$
In more detail, consider the projection map $\pi \colon V \to \mathcal{P}_2$, $\pi \colon (p,u) \mapsto p$. The first isomorphism theorem then says $\mathcal{P}_2 \cong V / \ker \pi$. But here, $\ker \pi = U'$.