In part $ii) $the part underlined in green suggests that we substitute an equation we get from when $v'=0$ to garner a solution of $s'$ for all time from the time when $v'=0$. However $v'$ does not remain $0$ for all time from the the time when $v'=0$, so why can we substitute the expression $v=\frac{s}{s+2}$ into the $s'$ equation?
Excellent question. I understand that you're confused by the phrasing of the answer, because the approach seems contradictory: first, you assume that $v$ is constant, from $v'=0$ you deduce that $v = \frac{s}{s+2}$, and substitute that into the dynamical equation for $s$ -- which leads to a nonconstant solution $s(t)$, which implies that $v(t) = \frac{s(t)}{s(t)+2}$ is nonconstant as well, contradicting the first assumption!
So, what's going on here? The clue is in the first sentence of the answer, that is, that $v$ changes much more rapidly than $s$. To quantify this, we need to assume that the only parameter in the problem, $e_0$, is very small, i.e. $0< e_0 \ll 1$. So, we have the system \begin{align} \frac{\text{d} s}{\text{d} t} &= e_0 \left[v - s(1-v)\right],\\ \frac{\text{d} v}{\text{d} t} &= s(1-v) -2v. \tag{1} \end{align} Since $e_0$ is very small, we can approximate the behaviour of $(1)$ very well by setting $e_0 = 0$, which leads to the so-called reduced system \begin{align} \frac{\text{d} s}{\text{d} t} &= 0,\\ \frac{\text{d} v}{\text{d} t} &= s(1-v) -2v. \tag{1a} \end{align} So, to leading order, we see that we can treat $s$ as constant (because it is changing really really slowly), and we're left with just the equation for $v$ $(1a)$. You can quite easily show that $v$ converges to its equilibrium value, for example by looking at its solution \begin{equation} v(t) = \frac{s}{s+2} + c_0\,e^{-(s+2)t}. \end{equation} Since $s>0$, we see that $v$ converges exponentially fast (with exponential rate $s+2$) to the stationary solution $v = \frac{s}{s+2}$.
Now, the above all happens pretty fast. What happens on the long run? I mean, we argued that $s$ is changing very slowly, but that is not exactly the same as $s$ being constant. To study the long time behaviour of the system $(1)$, we introduce the slow time variable $\tau = e_0 t$. In this slow time variable, the system $(1)$ is transformed to \begin{align} \frac{\text{d} s}{\text{d} \tau} &= \left[v - s(1-v)\right],\\ e_0 \frac{\text{d} v}{\text{d} \tau} &= s(1-v) -2v. \tag{2} \end{align} Again, we remember that $e_0$ is very small, so we can approximate the behaviour of $(2)$ very well by setting $e_0 = 0$, which gives the reduced system \begin{align} \frac{\text{d} s}{\text{d} \tau} &= \left[v - s(1-v)\right],\\ 0 &= s(1-v) -2v. \tag{2a} \end{align} As we see from the second equation, $v = \frac{s}{s+2}$. This means that on this slow time scale, $v$ is following the slow evolution of $s$.
So, combining the above descriptions, we can conclude that $v$ first converges very fast (in the fast time scale $t$) to the equilibrium value $v = \frac{s}{s+2}$, while $s$ hasn't notably changed during this convergence. Since the equilibrium $\frac{s}{s+2}$ is stable, $v$ will stay at, or very close to, its equilibrium value $\frac{s}{s+2}$, even if that value changes slowly in time (in the slow time variable $\tau$). So, because $\frac{s}{s+2}$ is changing very slowly, $v$ has ample time to 'catch up' with its equilibrium value $\frac{s}{s+2}$ again, because it converges exponentially fast to that value in the fast time $t$.
The approach sketched above is called multiple time scale analysis. For more information, in particular on the method of geometric singular perturbation theory, I can suggest
C. Kuehn, Multiple Time Scale Dynamics, Springer, 2015.