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). $\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\St}{St}$
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 here.
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)
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.
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 $(i,j)$.
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 $\HH_2(\SL(\Field);\Z) = 0$.