I've been corresponding via email with the OP about this, and he asked me to post an answer summarizing what I told him.  I apologize for the length of this answer -- this is really quite a long story.  I also apologize for sometimes butchering people's names.  I am copying this answer from a long document I sent the OP, and unfortunately autocorrect screws things up without telling me (eg it wants to call Steinberg "Sternberg", though I hope caught most of those occurrences).
$\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}}$ 



For $\Sp_{2g}(\Z/2)$, the correct theorem is that $\HH_2(\Sp_{2g}(\Z/2)) = 0$ for $g \geq 4$.  This should be attributed to Steinberg and is contained in the paper cited by the OP.  More generally, Steinberg showed that a similar theorem holds for $\Sp_{2g}(\Field_q)$.  What Stein did in the cited paper was show how to extend what Steinberg did to $\Sp_{2g}(\Z/k)$ where $k$ is not prime.  They both in fact dealt not just with the symplectic group, but also with more general finite Chevalley groups.

Another good reference for Steinberg's work is sections 6 and 7 of Steinberg's Yale lecture notes, which were never published but which are available <a href="http://www.ms.unimelb.edu.au/\~{}ram/Resources/YaleNotes.pdf">here</a>.

It is not easy to extract the above homological statement from Steinberg and Stein's papers since they are written in the language of algebraic k-theory.  What they are imitating is the calculation of $\HH_2(\SL_n(\Field_q))$ that is described in Milnor's book on algebraic k-theory, which I highly recommend reading.

To help you understand these papers, below I have written a guide to the calculation of $\HH_2(\SL_n(\Field_q))$ from Milnor's book.
(nb: since this is long, I'm going to post it in stages so that my work isn't lost if things crash)

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I will begin by recalling the theory of universal central extensions.  Let $G$ be a group.  A **central extension** of $G$ is a group $\tG$ together with a short exact sequence
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
such that $C$ is contained in the center of $\tG$.  This central extension is a **universal central extension**
if for any other central extension
$$1 \longrightarrow C' \longrightarrow \tG' \longrightarrow G \longrightarrow 1,$$
there exists a unique homomorphism $\tG \rightarrow \tG'$ such that the diagram
$\require{AMScd}$
$$\begin{CD}
1 @>>> C @>>> \tG @>>> G @>>> 1 \\
@.     @VVV   @VVV     @VV{=}V   @. \\
1 @>>> C' @>>> \tG' @>>> G @>>> 1
\end{CD}$$
commutes.  The usual argument shows that universal central extensions are unique if they exist, but they might
not exist.  The following theorem summarizes their properties.  A proof of it can be found in Theorem 5.7 and Corollary 5.8 of Milnor's book

**Theorem 1:** Let $G$ be a group.  Then $G$ has a universal central extension
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
if and only if $\HH_1(G;\Z) = 0$, in which case we have $C \cong \HH_2(G;\Z)$.

For perfect groups, this reduces the computation of $\HH_2(G;\Z)$ to the construction of the universal central
extension of $G$.

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