Deriving the Euler Equation I want to derive the Euler Equation for the following: 

$$max \sum\limits_{t=0}^{T} = \beta^{t}U(C_t)$$

$$s.t. C_t+K_{t+1} \leq f(K_t) , t=0,1,2,...,T-1$$ 
$$-K_{T+1} \leq 0$$

I'm a bit confused about why the F.O.C. have that: 

$$\frac{d\mathcal{L}}{dK_{t+1}}=-\lambda_t+\lambda_{t+1}f'(k_{t+1})$$

and how we combine the F.O.C to yield the Euler equation: 

$$U'(C_t)= \beta U'(C_{t+1})f'(k_{t+1})$$ 

I assume the F.O.C w.r.t. $K_{t+1}$ is such because of the inclusion of the intensive form of the production function but I am not exactly sure how and I really want to understand this completely.  I also need to make sure I understand how we are using the FOC to produce the Euler Equation. 
Can anyone provide a bit of clarity?
The question is quite straightforward, and you do not need the first
step that you have. You have (for some reason) a different multiplier
here for each time period. That is not the case- you simply have a
no-ponzi scheme condition, aka a transversality constraint. The problem
is expressed as $max\,\sum\beta^{t}U(c_{t})$ st $c_{t}+k_{t+1}\leq f(k_{t})$.
The stream of utility to the agent is $U=\sum\beta^{t}U(c_{t})=U(c_{0})+....\beta^{t}U(c_{t})+\beta^{t+1}U(c_{t+1})+...+\beta^{T}U(c_{T}).$

Substitute in $f(k_{t})-k_{t+1}$ for $c_{t}$. Then, we derive wrt
$k_{t+1}$ and set equal to 0:$\frac{\partial U}{\partial k_{t+1}}=\beta^{t}U'(c_{t})-\beta^{t+1}U'(c_{t+1})f'(k_{t+1})=0\Rightarrow U'(c_{t})=\beta U'(c_{t+1})f'(k_{t+1}).$

You do not need the other equation you referred to.