I've been corresponding via email with the OP about this (it is a paper of mine that she got these citations from), and she asked me to post an answer summarizing what I told her. 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).
$\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\St}{St}$
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.
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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 $G$.
-----------
I now describe what happens in "infinite rank". Let $R$ be a ring. Define
$$\GL(R) = \bigcup_{n=1}^{\infty} \GL_n(R)$$
and let $E(R)$ be the subgroup of $\GL(R)$ generated by elementary matrices.
The low-dimensional homology groups of $\GL(R)$ and $E(R)$
are closely connected to the algebraic K-theory of $R$. In particular,
$K_1(R)$ is by definition equal to $\HH_1(\GL(R);\Z)$.
It also turns out that $K_2(R) = \HH_2(E(R);\Z)$, but this
is a theorem rather than a definition. To understand what happens in finite rank, we'll have to understand
where this comes from.
There a group $\St(R)$ called the **Steinberg group**.
It is defined via generators and relations, and informally can be described as the group generated by
elementary matrices in $\GL(R)$ with the ``obvious relations'' between elementary matrices. Its precise
definition is as follows.
1. The generators are symbols $e_{ij}(r)$ with $i,j \geq 1$ distinct
and $r \in R$. This corresponds to the elementary matrix obtained from the identity matrix by putting an $r$ in
position $(i,j)$.
2. The relations are as follows.
- For distinct $i,j \geq 1$ and $r,r' \in R$, we have $e_{ij}(r)
\cdot e_{ij}(r') = e_{ij}(r+r')$.
- For distinct $i,j,k \geq 1$ and $r,r' \in R$, we have
$[e_{ij}(r),e_{jk}(r')] = e_{ik}(r r')$.
- For distinct $i,j \geq 1$ and distinct $k,l \geq 1$ such that $j
\neq k$ and $i \neq l$, we have $[e_{ij}(r),e_{kl}(r')] = 1$ for all
$r,r' \in R$.
Since all these relations hold in $\GL(R)$, there is a group homomorphism $\St(R) \rightarrow \GL(R)$ whose
image is $E(R)$. By definition, $K_2(R)$ is the kernel of this homomorphism, so we have a short
exact sequence
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1.$$
The main theorem concerning the Steinberg group is as follows (see Theorem 5.10 of Milnor's book).
**Theorem 2:** Let $R$ be a ring. Then the extension
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1$$
is the universal central extension of $E(R)$. In particular, $K_2(R)$ is an abelian group
and $\HH_2(E(R);\Z) = K_2(R)$.
This allows many concrete calculations. One very simple one is as follows.
**Example:** Since $\SL(\Z)$ is generated by elementary matrices, we have $E(\Z) = \SL(\Z)$. The group
$\HH_2(\SL(\Z);\Z)$ is then cyclic of order $2$. Identifying it with $K_2(\Z)$, the generator
is $(e_{12}(1) e_{21}(1)^{-1} e_{12}(1))^4$. Here the matrix
$$e_{12}(1) e_{21}(1) e_{12}(1) = \left(\begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix}\right)$$
is the one that rotates the plane by $90$ degrees, and hence has order $4$.
Assume now that $R$ is a commutative ring and let $u,v \in R^{\ast}$ be units. Define
$$D_u = \left(\begin{matrix}
u & 0 & 0 \\
0 & u^{-1} & 0 \\
0 & 0 & 1
\end{matrix}\right)$$
and
$$D_v' = \left(\begin{matrix}
v & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & v^{-1}
\end{matrix}\right).$$
It is not hard to see that diagonal matrices like these can be written as products of elementary matrices, so
$D_u, D_v' \in E(R)$. What is more, the matrices $D_u$ and $D_v'$ commute. Letting $\tD_u$ and $\tD_v'$ be
lifts of $D_u$ and $D_v'$ to $\St(R)$, we obtain an element
$[\tD_u,\tD_v'] \in K_2(R)$, which will be called a **symbol** and denoted $\{u,v\}$. It is not hard to see
that this does not depend on the choice of lifts.
The following theorem says that for fields, the symbols generate $K_2(R)$; see Corollary 9.13 of Milnor's book.
**Theorem 3:** Let $\Field$ be a field. Then $K_2(\Field)$ is generated by the set of symbols $\{u,v\}$ as $u$ and $v$
range over $\Field^{\ast}$.
For finite fields, the final piece of the puzzle is as follows; see Corollary 9.9 of Milnor's book.
**Theorem 4:** Let $\Field$ be a finite field. Then $\{u,v\} = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above with the fact that $\SL(\Field)$ is generated by elementary matrices for a field $\Field$,
we deduce the following theorem.
**Theorem 5:** Let $\Field$ be a finite field. Then $\HH_2(\SL(\Field);\Z) = 0$.
-------------
Of course, what we are really interested in is $\GL_n(R)$, not $\GL(R)$. Define $E_n(R)$ and $\St_n(R)$ in the obvious
way. There is still a surjection $\St_n(R) \rightarrow E_n(R)$; denote its kernel by $C_n(R)$, so we have a short
exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1.$$
Associated to the natural inclusion $\GL_n(R) \hookrightarrow \GL_{n+1}(R)$ used to define $\GL(R)$ are an inclusion
$E_n(R) \hookrightarrow E_{n+1}(R)$ and homomorphisms $\St_n(R) \rightarrow \St_{n+1}(R)$ and $C_n(R) \rightarrow C_{n+1}(R)$
that fit into a commutative diagram
$$\begin{CD}
1 @>>> C_n(R) @>>> \St_n(R) @>>> E_n(R) @>>> 1 \\
@. @VVV @VVV @VVV @. \\
1 @>>> C_{n+1}(R) @>>> \St_{n+1}(R) @>>> E_{n+1}(R) @>>> 1.
\end{CD}$$
It is clear that $\St_n(R)$ is the limit of
$$\St_1(R) \rightarrow \St_2(R) \rightarrow \St_3(R) \rightarrow \cdots$$
and that $K_2(R)$ is the limit of
$$C_1(R) \rightarrow C_2(R) \rightarrow C_3(R) \rightarrow \cdots.$$
In the ``ideal'' situation, we would have theorems of the following form.
1. For $n$ sufficiently large, the exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$; and hence $C_n(R)$ is an abelian group and $\HH_2(E_n(R);\Z) = C_n(R)$.
2. For $n$ sufficiently large, the map $C_n(R) \rightarrow C_n(R)$ is surjective (this is called **surjective stability**).
3. For $n$ sufficiently large, the map $C_n(R) \rightarrow C_n(R)$ is injective (this is called **injective stability**).
If these conditions are satisfied, then we would have $\HH_2(E_n(R);\Z) = K_2(R)$ for $n$ sufficiently large.
Unfortunately, one can give examples where these fail. However, they do hold for many rings; in particular, they
hold for fields. Our goal is
to talk about finite fields, so we will not try to give particularly general statements. We begin with the
bit about being a universal central extension. The proof of Theorem 2 can
be followed to deduce the following.
**Theorem 6:** Let $R$ be a ring and let $n \geq 5$. Assume that $C_n(R)$ is contained in the center of $\St_n(R)$. Then the extension
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$. In particular,
$\HH_2(E_n(R);\Z) = C_n(R)$.
As the following theorem shows, the condition in this theorem is satisfied for fields; see Theorem 9.12 of Milnor's book.
**Theorem 7:** Let $\Field$ be a field. Then $C_n(\Field)$ is contained in the center of $\St_n(\Field)$ for $n \geq 3$, and hence
$\HH_2(\SL_n(\Field);\Z) = C_n(\Field)$ for $n \geq 5$
**Remark:** In fact, it turns out that
$$1 \longrightarrow C_n(\Field) \longrightarrow \St_n(\Field) \longrightarrow \SL_n(\Field) \longrightarrow 1$$
is the universal central extension of $\SL_n(\Field)$ for $n \geq 3$ except for the following exceptions:
1. $n=3$ and $\Field = \Field_2$, and
2. $n=3$ and $\Field = \Field_4$, and
3. $n=4$ and $\Field = \Field_2$.
This was proved by Steinberg.
We now turn to injective and surjective stability. The key is the notion of symbol. If $R$ is a commutative ring
and $u,v \in R^{\ast}$ are units, then we can define a symbol $\{u,v\}_n \in C_n(R)$ for any $n \geq 3$ in the obvious
way. We then have the following; see Theorem 9.11 of Milnor's book.
**Theorem 8:** Let $R$ be a commutative ring and let $n \geq 3$. Assume that $C_n(R)$ is contained in the center of $\St_n(R)$.
Then $C_n(R)$ is generated by the set of symbols $\{u,v\}_n$ as $u$ and $v$ range over the units of $R$.
**Remark:** Theorem 7 and Theorem 8 combine to show that if $\Field$
is a field, then $K_2(\Field)$ is generated by symbols $\{u,v\}$ for $u,v \in \Field^{\ast}$. This is
precisely Theorem 3 above, and in fact this is how Theorem
3 is proved.
The proof of Theorem 4 above also works for the symbols $\{u,v\}_n$ and
gives the following.
**Theorem 9:** Let $\Field$ be a finite field and $n \geq 3$. Then $\{u,v\}_n = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above, we deduce the following.
**Theorem 10:** Let $\Field$ be a finite field. Then $\HH_2(\SL_n(\Field);\Z) = 0$ for $n \geq 5$.
**Remark:** In fact, as in the remark after Theorem 7, one can show that
$\HH_2(\SL_n(\Field);\Z) = 0$ for $\Field$ a finite field and $n \geq 3$ except for the following exceptions:
1. $n=3$ and $\Field = \Field_2$, and
2. $n=3$ and $\Field = \Field_4$, and
3. $n=4$ and $\Field = \Field_2$.