Do discrete embeddings of surface groups not necessarily carry an embedding of SL_2? We can get arithmetic lattices isomorphic to free groups in $\mathrm{SL}_2\mathbb{R}$, so in general we can’t expect homomorphisms of lattices into semisimple Lie groups to say much about $\mathrm{SL}_2\mathbb{R}$, even if we require the image to be discrete. In particular, unlike in higher-rank cases, such homomorphisms won’t necessarily extend to homomorphisms of $\mathrm{SL}_2\mathbb{R}$.
However, if $\Gamma<\mathrm{SL}_2\mathbb{R}$ is cocompact and we have an discrete embedding (by which I mean a map which is both a homeomorphism and an isomorphism onto its image) of $\Gamma$ into some semisimple Lie group $G$, will we not necessarily get a copy of $\mathrm{SL}_2\mathbb{R}$ in $G$ containing (the image of) $\Gamma$?
By the considerations above, the image of $\Gamma$ in such a copy of $\mathrm{SL}_2\mathbb{R}$ might not look like it originally did (meaning the embedding of $\Gamma$ hasn’t necessarily extended to all of $\mathrm{SL}_2\mathbb{R}$), but part of me imagines that the image of $\Gamma$ should still determine something. Is this naïve?
 A: Homomorphisms from surface groups are very flexible, so there are indeed such examples.
For instance, use: Breuillard–Gelander–Souto–Storm (Dense embeddings of surface groups.
Geom. Topol. 10 (2006), 1373–1389) (DOI link)
They proved, for a surface group $\Gamma$ ($\pi_1$ of oriented closed surface of genus $\ge 2$) that, among others, $\Gamma$ embeds densely into arbitrary nontrivial connected semisimple Lie groups.
To deduce a discrete embedding, argue as follows: choose a dense embedding  into $G$ (with $G$ pretty arbitrary, even $G=\mathrm{SL}_2(\mathbf{R})$ is fine) and consider a standard discrete embedding into $\mathrm{PSL}_2(\mathbf{R})$. Then the diagonal embedding into $G\times\mathrm{SL}_2(\mathbf{R})$ has a Zariski-dense image, so its image is not contained in any copy of any subgroup locally isomorphic to $\mathrm{SL}_2(\mathbf{R})$.

Variant, not using the above reference [BGSS]:
Choose two Fuchsian embeddings $i_1,i_2:\Gamma\to H$ where $H=\mathrm{PSL}_2(\mathbf{R})$, such that the images of $i_1$ and $i_2$ are not conjugate under $\mathrm{PGL}_2(\mathbf{R})$. Then the diagonal embedding $i_1\times i_2:\Gamma\to H\times H$ works (its image is Zariski-dense in $H^2$).
