For your first question, here's a simple proof. 

1. Observe that $\partial_{ii}^2 e_k = 0$ for any $i\in \{1,\ldots,n\}$. This implies 
$$ \partial^2_{ii} (\sigma^k f) = 0 \tag{1}$$
for any $i$, where $\sigma = \sum x_i$. 
2. Observe that 
$$ \partial_i \sigma = 1 $$
for any $i$, and hence 
$$ \partial_i \sigma^k = k \sigma^{k-1} \tag{2}$$
by the chain rule. 
3. The product rule applied to (1) using (2) gives
$$ 0 = k(k-1)\sigma^{k-2} f + 2 k \sigma^{k-1} \partial_i f + \sigma^k \partial^2_{ii} f$$
from which we divide by $\sigma^k$ to obtain
$$ \partial^2_{ii} f + k(k-1)\sigma^{-2} f + 2 k \sigma^{-1} \partial_i f = 0 \tag{3}$$
Recall that this holds _individually for any $i$_. 
4. Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally
$$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
5. Now observe that $f$ is, by definition, a [homogeneous function of degree 0](https://en.wikipedia.org/wiki/Homogeneous_function). So the homogeneous derivative $\sum x_i \partial_i f = 0$. This kills the last term in \tag{4). The remainder is exactly what you wanted to show.