The OP's confusion comes from the fact that he or she does not apply the chain rule correctly and interprets the Euler condition
$u'(c_1) = \beta (1+r) u'(c_2)$
as
$ u_{C_1}(C_1) = \beta(1+r) u_{\boldsymbol{C_1}}(C_2) $
This is not the correct way to read the Euler condition and one should really read
$ u_{C_1}(C_1) = \beta(1+r) u_{\boldsymbol{C_2}}(C_2) $
Any doubt can be cleared by rederiving the FOC. The maximisation problem is
$\max_{C_1,C_2} u(C_1) + \beta u(C_2) \qquad s.t. \qquad $
\begin{equation*}
C_1 + \frac{C_2}{1+r} = Y_1 + \frac{Y_2}{1+r} \implies C_2 = {-C_1}({1+r}) + Y_1(1+r) + {Y_2}
\end{equation*}
So substituting the constraint, the problem can be written as
\begin{align}\max_{C_1} u\Big(C_1\Big) + \beta u\Big(\underbrace{{-C_1}({1+r}) + Y_1(1+r) + {Y_2}}_{=C_2}\Big)\end{align}
To find the FOC, one needs to apply the chain rule to the second period's utility function. Applying the chain rule carefully we have
\begin{align} \frac{\partial \beta u(C_2)}{\partial C_1} = \beta \frac{\partial u(C_2)}{\partial C_2} \frac{\partial C_2}{\partial C_1}\end{align}
In our case that is
\begin{align} \frac{\partial \beta u(C_2)}{\partial C_1} = \beta \frac{\partial u(C_2)}{\partial C_2} (-1)(1+r)\end{align}
Or to put it in yet another way
\begin{align}\frac{\partial \beta u(C_2)}{\partial C_1} & = \frac{\partial \big[\beta u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial C_1}\\ & = \beta \frac{\partial \big[\ u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)} \frac{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)}{\partial C_1} \\ & = \beta \frac{\partial \big[\ u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)} (-1) (1+r) \\ &= \beta \frac{\partial u(C_2)}{\partial C_2} (-1)(1+r) \end{align}
Then, by definition, given a utility function $U(c_1,c_2)$, the MRS of good one with respect to good two is
$MRS := \frac{{\partial U}/{\partial c_1}}{{\partial U}/{\partial c_2}}$
In your case, as stated in your notes, you get
$MRS = \frac{{\partial u}/{\partial c_1}}{\beta{\partial u}/{\partial c_2}} = \frac{u'(c_1)}{\beta{u'(c_2)}}$
At an equilibrium, when consumers maximize utility, the Euler condition must be satisfied, that is
$u'(c_1) = \beta (1+r) u'(c_2)$
So replace this specific value for $u'(c_1)$ in the formula for the MRS and you get
$MRS = \frac{u'(c_1)}{\beta{u'(c_2)}} = \frac{\beta (1+r) u'(c_2)}{\beta{u'(c_2)}} = (1+r)$.
As a good exercise, try to convince yourself that if the MRS takes another value, then the consumer can benefit from reallocating resources between the first and the second period.
So far so good. Next you make the derivative of $\mathcal L(c,l,\lambda)$ w.r.t $\lambda$, which is just the constraint.
$$\frac{\partial\mathcal L(c,l,\lambda)}{\partial \lambda}=24w-pc-wl=0$$
In combination with $l^*= 24-\left(\frac{w}{p}\right)^{\eta}$ we obtain
$$24w-pc-w\cdot \left(24-\left(\frac{w}{p}\right)^{\eta} \right)=0$$
It remains to solve the equation for $c$. I don´t think that you need the next steps. So I hide them and you can compare them with your own steps at the end.
$$-pc+w\cdot \left(\frac{w}{p}\right)^{\eta} =0$$
$$c^*=\frac{w}p\cdot \left(\frac{w}{p}\right)^{\eta} $$
$$c^*= \left(\frac{w}{p}\right)^{\eta+1} $$
Best Answer
The cost is an increasing function of $c_1$, so we can take $c_1 = w-s$.
If $\beta<0$, then we can choose $c_2$ close to zero to get arbitrarily large cost, so presumably we have $\beta \ge 0$. In this case, the cost is a non-decreasing function of $c_2$, so we should take $c_2 = Rs$. The utility now reduces to the unconstrained $f(s)= \log (w-s) + \beta \log (Rs)$.
Differentiating gives $-\frac{1}{w-s} + \beta \frac{1}{s} = 0$ (note the first minus sign!), which yields $s = \frac{\beta}{1+\beta} w$.