Let $\mathcal{L}$ be a language containing a binary relation symbol $R$. I need to prove that if $T$ is a universal $\mathcal{L}$-theory (by which I mean that $T$ is a collection of universal formulas) such that $T\vdash\forall x\exists yR(x,y)$ then for some $\mathcal{L}$-terms $t_1(x),\ldots,t_n(x)$ we have $T\vdash\forall x\bigvee_{i\leq n}R(x,t_i(x))$.
I only know that if $T$ is universal then the collection of models of $T$ is closed under submodels. Therefore if $\mathcal{M}\vDash T$ and $m\in \mathcal{M}$ then $\{t(m)\mid t\,\mathcal{L}$-term$\}$ is the universe of a substructure of $\mathcal{M}$ and this implies $\mathcal{M}\vDash R(m,t(m))$ for some $\mathcal{L}$-term $t$. I don't know if this is correct or useful at all, if so I believe I should continue with some compactness argument but I don't know how.
You're exactly right that you need to use a compactness argument, and you've just about set it up!
Add a constant $c$ to the language, and consider the theory $T \cup \{\lnot R(c,t(c))\mid t\text{ an }\mathcal{L}\text{-term}\}$. You've observed that the theory is inconsistent. So some finite subset is inconsistent, i.e. $T\models \lnot (\bigwedge_{i=1}^n\lnot R(c,t_i(c)))$ for some finite set of $\mathcal{L}$-terms $\{t_1,\dots,t_n\}$.
Now observe that since $c$ doesn't appear on the left hand side of the $\models$, you can replace it with a universally quantified variable.