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.
Hint: First, start off with the constrained maximization problem and then use the Lagrangian method to get the first-order conditions. Then use the Kuhn-Tucker method (see Mathematics for Economist by Simon and Blume) to solve for your optimal demands.
The consumer solves the following maximization problem:
\begin{align*}
\max_{q_1,q_2} \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) \\
\text{s.t. } p_1 q_1 + p_2q_2 \le m.
\end{align*}
The Lagrangian associated with maximization problem is
\begin{equation*}
\mathcal{L} = \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) + \lambda[m - p_1 q_1 - p_2q_2]
\end{equation*}
Then take the derivatives of $\mathcal{L}$ wrt to $q_{1}$, $q_{2}$, and $\lambda$ to get the following first-order conditions:
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial \lambda}=m-p_{1}q_{1}-p_{2}q_{2}\geq0
\end{equation}
Then we use the Kuhn-Tucker method:
$\lambda[m-p_{1}q_{1}-p_{2}q_{2}] = 0$ is the complementary slackness condition on $\lambda$
and
$\lambda\geq0$ is the non-negativity constraint on $\lambda$.
We can have two cases: either $\lambda>0$ or $\lambda=0$. I show what happens for $\lambda>0$ and then you can check what happens for $\lambda=0$.
If $\lambda>0$ then $m-p_{1}q_{1}-p_{2}q_{2}=0$ by the complementary slackness condition. In other words, this is the case where the budget constraint is binding, that is, $m=p_{1}q_{1}-p_{2}q_{2}$. Then we check whether the first-order conditions and the non-negativity condition hold.
After doing so we are left with the following set of equations:
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0
\end{equation}
\begin{equation}
m-p_{1}q_{1}-m_{2}q_{2}=0
\end{equation}
Then substitute away $\lambda$ from the first two equations and solve for $q_{1}$ and $q_{2}$ using the binding budget constraint.
Edit: For the unconstrained problem you have no budget constraint so your set up is conceptually wrong. The way you get an unconstrained problem from a constrained problem is that you substitute away your budget constraint in the utility function. Also I haven't imposed non-negative conditions on $q_{1}$ and $q_{2}$, as then the process would be quite tedious.
Best Answer
The consumer maximizes intertemporal utility over the whole stream of consumption (and I suspect income is exogenous here). Also for the result to pass through, the discount rate for utility must be identical with the discount rate for consumption-income (usually they aren't, the first being related to pure time preference, the second to interest rates). But assume they are identical. Then we have
\begin{equation*} \max_{\{c_{t+j}\}|_{j=0}^{\infty}} \sum_{j=0}^{\infty}E_t[b^ju(c_{t+j})] \quad \textrm{s.t.}\quad \sum_{j=0}^\infty E_t[\beta^jc_{t+j}] = \sum_{j=0}^\infty E_t [\beta^j y_{t+j}] \end{equation*} Since the budget constraint is written in present-value form, then we have one lagrange multiplier for all periods, so the Langrangean is
\begin{equation*} L =\sum_{j=0}^{\infty}E_t[\beta^ju(c_{t+j})] + \lambda\left[ \sum_{j=0}^\infty E_t[\beta^jc_{t+j}] - \sum_{j=0}^\infty E_t [\beta^j y_{t+j}] \right] \end{equation*}
The consumer solves this problem for $j=0,1,...$. For $j=0$ the expectations operator goes away and the first order condition is
$$j=0\qquad u'(c_t) + \lambda = 0 \Rightarrow a - bc_t = \lambda \Rightarrow c_t = \frac {a+\lambda}{b}$$
For $j=k$ we have
$$j=k\qquad E_t[\beta^ku'(c_{t+k})] + \lambda\beta^k = 0 \Rightarrow \beta^k\left(a - bE_t[c_{t+k}]\right) = \lambda\beta^k \Rightarrow E_t[c_{t+k}] = \frac {a+\lambda}{b}$$
So
$$E_t[c_{t+k}] = c_t$$
Intuitively, this happens because essentially the consumer solves a static problem, although in an intertemporal guise. Decision-making becomes truly dynamic when there exist factors that can be accumulated, creating the trade-off between present and future.