The statement of the second m.v.t. I use to know concerns continuous functions, so its proof rely on the first m.v.t. for continuous functions. Hence I had to work out a new proof... And, quite surprisingly, it seems that both $G(a),G(b)$ and $G(a^+),G(b^-)$ work in the equality.
Therefore, maybe there is something wrong in my proof... You have to check carefully. ^^
Proof: Let $G,\phi:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable over $[a,b]$ and $G(x)$ be decreasing. It is obvious that it suffices to prove that there exists $\xi ,\eta \in [a,b]$ s.t.:
$\int_a^b \phi G =G(a^+)\int_a^\xi \phi =G(a)\int_a^\eta \phi$
for nonnegative $G(x)$, because if $G(x)$ takes both positive and negative values, then $G(x)-G(b^-)$ and $G(x)-G(b)$ are nonnegative (the former is negative in only $b$, but a point is negligible w.r.t. integration).
Set:
$l_k:=\inf_{[x_k,x_{k+1}]} \phi,\ L_k :=\sup_{[x_k,x_{k+1}]} \phi$
$m_k:=\inf_{[x_k,x_{k+1}]} G,\ M_k :=\sup_{[x_k,x_{k+1}]} G$
and note that the range of $\phi(x) G(x)\Big|_{[x_k,x_{k+1}]}$ is $\subseteq [\min \{ l_k m_k,l_k M_k\}, \max \{ L_k m_k, L_k M_k\}]=:I_k$. Let $D=\{a=x_0 be a partition of $[a,b]$ and write the Riemann sum:
$\sigma_D (\lambda_0\mu_0,\ldots ,\lambda_n \mu_n) :=\sum_{k=0}^n \lambda_k \mu_k (x_{k+1}-x_k)$,
where $\lambda_k\in [l_k,L_k],\ \mu_k \in [m_k, M_k]$ (so that $\lambda_k\mu_k \in I_k$): it is well known that $\sigma_D(\lambda_k\mu_k) \to \int_a^b \phi G$ when the diameter of $D$ shrinks to $0$ independently on how we choose $\lambda_k, \mu_k$.
Now set $\Phi (x):=\int_a^x \phi$ and choose $\lambda_k$ s.t. $\lambda_k (x_{k+1}-x_k)=\int_{x_k}^{x_{k+1}}\phi =\Phi(x_{k+1}) -\Phi (x_k)$ (which is possible for the first m.v.t.): therefore the Riemann sum rewrites:
$\sigma_D(\lambda_k\mu_k)=\sum_{k=0}^n \mu_k [\Phi (x_{k+1})-\Phi (x_k)] = \sum_{k=0}^{n-1} (\mu_k -\mu_{k+1})\Phi (x_{k+1}) + \mu_{n} \Phi (x_{n+1})$.
$\Phi(x)$ is a bounded function in $[a,b]$ (for $\phi$ is), and the differences $\mu_-\mu_{k+1}$ are nonnegative (because $G(x)$ decreases one has $m_k\geq M_{k+1}$, and $\mu_k \in [m_k,M_k], \mu_{k+1}\in [m_{k+1},M_{k+1}]$), therefore:
(*) $p\mu_0 \leq \sigma_D(\lambda_k \mu_k)\leq P\mu_0$,
where $p:=\inf_{[a,b]} \Phi$ and $P:=\sup_{[a,b]} \Phi$.
Now it is clear that in (*) one can choose $\mu_0$ arbitrarily: in particular, one can choose $\mu_0=G(a^+)$ or $\mu_0=G(a)$, and in each case, one gets an uniform bound (w.r.t. $D$) for $\sigma_D(\lambda_k,\mu_k)$.
Finally, letting the diameter of $D$ tend to $0^+$, one has:
$p\mu_0\leq \int_a^b\phi G \leq P\mu_0$,
hence applying the first m.v.t. to the continuous function $\Phi(x)$ one has the claim, with two different point $\xi,\eta$ (in general) depending on the choice of $\mu_0$.