This is almost never true, only in the most trivial cases, and conditions to make it true will be contrived.
Here is a very simple example: suppose that $k = 1$, $n = 2$, that $X_1 = 1$ with probability 1 and that $X_2$ is a standard Bernoulli random variable. (That is, takes values 0 and 1 with probability $\tfrac12$ each.) Then
$$
\Bbb E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right)= \Bbb E\left(\frac1{1+X_2}\right) = \frac34 \neq \frac23 = \frac1{\Bbb E\left(1+X_2\right)} = \frac{\sum_{j=1}^k E[X_j]}{\sum_{i=1}^n E[X_i]}.
$$
Edit: Below is an earlier version of this answer, where I incorrectly read the variables in numerator and denominator as distinct; it shows more broadly that the key problem is that taking the inverse and $\mathbb E$ just do not commute.
Note that your observation for the numerator is correct, and if random variables $X, Y$ are independent then $\mathbb E[XY] = \mathbb E[X]\mathbb E[Y]$. This means that in the (already special) case of independent variables, we furthermore have
$$
\sum_{j=1}^k \Bbb E\left[\frac{ X_j}{ \sum_{i=1}^n X_i}\right] = \sum_{j=1}^k \Bbb E[X_j] \Bbb E\left[\frac1{ \sum_{i=1}^n X_i}\right],
$$
and because you already know that you can distribute sums over $\Bbb E$ the interesting question, really, is whether or not
$$
\Bbb E\left[\frac1X\right] \overset{?}{=} \frac1{\Bbb E[X]},
$$
and it is easy to verify in simple cases that this does not hold. E.g., $X$ takes values $1, 2$ both with probability $\tfrac12$: then $\Bbb E[1/X] = \tfrac34$ while $1/\Bbb E[X] = \tfrac23$.