To elaborate on Dmitri Pavlov's comment, your questions have been answered in a recent paper by Yehonatan Sella. Sella give an example (Example 0.3) of a locally contractible space $X$ for which $\mathcal{S}^k(X) \xrightarrow{\mathrm{shf}_X} \tilde{\mathcal{S}}^k(X)$ is not surjective. Sella's space has $5$ points, and the argument that $\mathrm{shf}_X$ is not surjective heavily exploits the non-Hausdorff nature of the space. It remains unclear to me whether you can get a counterexample which is Hausdorff.
Here is a different counterexample from the one Sella gives (but based on the same idea) which I think more clearly demonstrates what is going on geometrically. Take $X=\mathbb{R}\cup\{0'\}$ to be the line with two origins, so neighborhoods of $0'$ look like neighborhoods of $0$ except with $0$ replaced by $0'$. Define $1$-cochains $\beta_0$ on $U=X\setminus\{0'\}$ and $\beta_1$ on $V=X\setminus\{0\}$ as follows. Define $\beta_0=0$. Define $\beta_1(\alpha)=0$ for any $1$-simplex $\alpha$ unless $\alpha$ is the linear path from $\epsilon$ to $2\epsilon$ for some $\epsilon>0$, in which case $\beta_1(\alpha)=1$. Note that any point in $U\cap V=\mathbb{R}\setminus\{0\}$ has a neighborhood which does not contain $[\epsilon,2\epsilon]$ for any $\epsilon>0$, and so $\beta_0$ and $\beta_1$ agree locally on $U\cap V$. However, there is no $1$-cochain $\beta$ on $X$ which agrees with $\beta_0$ in a neighborhood of $0$ and which also agrees with $\beta_1$ on a neighborhood of $0'$, since such neighborhoods would contain $[\epsilon,2\epsilon]$ for all sufficiently small $\epsilon$.
After showing this counterexample, Sella then gives an alternate proof that singular cohomology agrees with sheaf cohomology that only requires the space to be locally contractible (or actually even just semi-locally contractible, meaning that there is a basis of open sets $U$ such that the inclusion map $U\to X$ is nullhomotopic). The idea is that instead of taking the presheaf quotient $\mathcal{S}^k/\mathcal{S}^k_0$, which, as seen above, may not be a sheaf, you take a smaller presheaf quotient $\mathcal{C}^k$. This quotient $\mathcal{C}^k$ can be shown to be a sheaf, and still forms a flasque resolution of the constant sheaf and has the same cohomology on global sections as $\mathcal{S}^k$.
Here's the motivation behind the construction, if I'm understanding it correctly. The reason that it's hard to prove that $\mathcal{S}^k/\mathcal{S}^k_0$ is a sheaf is that the equivalence relation on cochains is too weak. Two cochains are equivalent iff they agree when restricted to some neighborhood of each point. However, the restriction to a neighborhood of a point still involves plenty of simplices that don't pass through that point, and so these "local" conditions at different points interact with each other and are hard to simultaneously satisfy at all points. This is what's going on with the counterexample above: you can't glue $\beta_0$ and $\beta_1$ because the constraints of being locally equal to $\beta_0$ at $0$ and locally equal to $\beta_1$ at $0'$ interfere with each other.
To fix this, Sella defines a stronger equivalence relation on cochains that makes the "local equality" at different points independent of each other. Slightly simplifying, he declares two cochains $\beta,\beta'\in\mathcal{S}^k(U)$ to be equivalent if for each $x\in U$ there is a neighborhood $V$ of $x$ such that $\beta(\alpha)=\beta'(\alpha)$ for all simplices $\alpha$ which are contained in $V$ and have barycenter $x$. If you left out the barycenter condition, this would just be equivalence mod $\mathcal{S}^k(U)_0$. Note that this barycenter condition means that to satisfy this local condition at a point $x$, you only care about simplices with barycenter $x$, which do not affect the local condition at any other point.
Sella then defines $\mathcal{C}^k(U)$ to be the quotient of $\mathcal{S}^k(U)$ by this equivalence relation (actually, he uses a more complicated equivalence relation in order to make this construction compatible with the coboundary operation on cochains, but never mind that). Because of the independence of the local conditions at different points, it is easy to show that $\mathcal{C}^k$ is a sheaf (to glue compatible sections, just define a cochain whose value on simplices with barycenter $x$ is given by the germ of one of your sections at $x$). It is then not hard to show that these sheaves form a flasque resolution of the constant sheaf. The hard part is to show that $\mathcal{C}^\bullet(X)$ still has the same cohomology as $\mathcal{S}^\bullet(X)$, and this is what occupies the bulk of Sella's paper.
(Disclaimer: I have not carefully read Sella's paper, and the above is just the impression of the main ideas which I got from a cursory reading.)
Best Answer
If you want $H^k(X, \mathbb{R})$ to reproduce what people normally mean by "singular cohomology with coefficients in $\mathbb{R}$" then you need to give $\mathbb{R}$ the discrete topology; with the Euclidean topology $\mathbb{R}$ is contractible so cohomology with coefficients in $\mathbb{R}$ is trivial.
In general, if $0 \to A \to B \to B/A \to 0$ is a short exact sequence of abelian groups, then we get a coefficient long exact sequence in cohomology
$$\dots \to H^k(X, A) \to H^k(X, B) \to H^k(X, B/A) \xrightarrow{\beta} H^{k+1}(X, A) \to \dots $$
where the connecting homomorphism $\beta$ is a Bockstein homomorphism. Applied to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$ (of discrete groups, so we also give $\mathbb{R}/\mathbb{Z}$ the discrete topology) this produces a long exact sequence
$$\dots \to H^k(X, \mathbb{Z}) \to H^k(X, \mathbb{R}) \to H^k(X, \mathbb{R}/\mathbb{Z}) \xrightarrow{\beta} H^{k+1}(X, \mathbb{Z}) \to \dots$$
showing that the map $H^k(X, \mathbb{R}) \to H^k(X, \mathbb{R}/\mathbb{Z})$ is neither injective nor surjective in general; it's not injective in general since by exactness its kernel is the image of $H^k(X, \mathbb{Z}) \to H^k(X, \mathbb{R})$ which is generally nonzero (if everything in sight is finitely generated it's exactly the torsion-free quotient of $H^k(X)$) and it's not surjective in general since by exactness its image is the kernel of the Bockstein homomorphism $\beta$, which is generally nonzero (since by exactness again the image of $\beta$ is the kernel of $H^{k+1}(X, \mathbb{Z}) \to H^{k+1}(X, \mathbb{R})$ which again if everything is finitely generated is exactly the torsion subgroup of $H^{k+1}(X)$).
If on the other hand you want $\mathbb{R}$ and $\mathbb{R}/\mathbb{Z}$ to have the Euclidean topology then we still get a long exact sequence as above except that the terms $H^k(X, \mathbb{R})$ vanish as above, and we conclude that in this case the Bockstein just produces an isomorphism $H^k(X, \mathbb{R}/\mathbb{Z}) \cong H^{k+1}(X, \mathbb{Z})$, corresponding to the delooping isomorphism $B^k \mathbb{R}/\mathbb{Z} \cong B^{k+1} \mathbb{Z}$ of classifying spaces.