This is only a partial answer, but probably too long to be a comment. One can count, for fixed $k$, the number of sums of $k$ consecutive primes.
Since we are concerned with consecutive primes, each sum is indexed by the smallest summand. For a prime $p$ let us write $p = p^{(1)}$, and $p^{(2)}, \cdots, p^{(k)}$ to be the next $k-1$ primes in order. We are thus considering the counting function
$$\displaystyle \pi^{(k)}(X) = \# \{p : p^{(1)} + \cdots + p^{(k)} \leq X \}.$$
Note that for fixed $k$ there is no problem with double counting, since the sums are strictly increasing.
We now use the fact that we can control the gaps between consecutive primes. Indeed, one can show that for sufficiently large $X$ the interval $[X, X + X^{0.525})$ contains a prime (this is a theorem due to Baker, Harman, and Pintz). Therefore, it follows that
$$\displaystyle p^{(1)} + \cdots + p^{(k)} = kp^{(1)} + O \left(p^{0.525} \right).$$
Hence, we see that
$$\displaystyle \pi^{(k)}(X) = \pi(X/k) + E(X),$$
where $E(X)$ counts the number of primes $p$ such that $kp \leq X$ but $p^{(1)} + \cdots + p^{(k)} > X$. This only happens for $p \in [X/k - O(X^{0.525}), X/k)$ and as long as $k$ is fixed, we can estimate the number of such primes by counting primes in short intervals, getting that there are $O(X^{0.525}/\log X)$ such primes. This error is far smaller than the error inherent in $\pi(X/k)$ when pulling out the main term $(X/k)/\log(X/k)$, so we can ignore it. Hence
$$\displaystyle \pi^{(k)}(X) \sim_k \frac{X}{k \log X}.$$
The arguments above give quite a bit of flexibility in terms of allowing $k$ to tend to infinity with $X$, but the exact range of uniformity probably takes a bit of work to obtain. The issue of repeat representations however will likely be a thornier issue.
