In my mind "the" example of a kink is
$F(t) = 0$ for $t \leq 0$
$F(t) = t$ for $t \geq 0$.
This is zero for a while and then suddenly starts to increase. It is also clearly the integral of
$f(t) = 0$ for $t \leq 0$
$f(t) = 1$ for $t > 0$.
This is a source of kinks in higher derivatives too, because you can easily take e.g.
$G(t) = 0$ for $t \leq 0$
$G(t) = t^3$ for $t \geq 0$.
Clearly continuous. And 1st derivative is continuous. Second derivative is continuous but with a kink.
Here's a partial answer. To begin, I'll prove the following result.
Lemma 1: Let $\sigma$ denote the signum function,
$$
\sigma(x) = \begin{cases}
-1 & x < 0\\
0 & x = 0\\
1 & x > 0.
\end{cases}
$$
We find that $S(\sigma)$ exists, with $S(\sigma) = \sigma$.
Proof outline: First, a quick note: it is easy to see that if $S(\sigma)$ exists, then it must hold that $S(\sigma)(0) = 0$. Indeed, the fact that $\sigma$ is an odd function ensures that each truncated Fourier series
$$
S_N(\sigma)(x) := \sum_{n=-N}^N \hat f(n)e^{inx}
$$
is itself odd. Thus, $S_N(\sigma)(0) = 0$ for all $N$, so that $\lim_{N \to \infty} S_N(\sigma)(0) = 0$.
As for proof of existence, it suffices to compute the Fourier series directly. One finds that
$$
S_N(f) = \frac 4{\pi} \sum_{n\text{ odd}, n \leq N} \frac 1n \sin\left( \frac{n x}{2}\right).
$$
At all $x$, this the above is the partial sum for a convergent alternating series. It remains to argue/show that this function converges to $\sigma(x)$ at the non-zero values of $x \in [-\pi,\pi]$.
I'll use the following facts without proof.
Lemma 2: Given functions $f,g$ with convergent Fourier series, $S(f + g) = S(f) + S(g)$
Lemma 3: If $f$ is continuous over $[-\pi,\pi]$ and differentiable except at finitely many points, then $S(f) = f$.
From there, the general result may be argued as follows. Given a suitable $f$ satisfying the hypothesis of the statement, we can build a piecewise constant function $g(x) = \sum_{j=1}^n a_j \sigma (x - x_j)$, where $x_1,\dots,x_n$ are the points at which $f$ is discontinuous and $a_j = \frac 12 (f(x_j^+) - f(x_j^-))$, such that $f - g$ is continuous. By combining the corollary of Lemma 1 and the other lemmas, one can deduce the more general result.
Best Answer
If $f$ has a primitive $F$, by definition $f$ satisfies $F'=f$. Now, by Darboux theorem $f$ satisfies the IVP.
A function with a jump discontinuity cannot satisfy the IVP:
First proof: analytical
let $\varepsilon=\frac{|f(x_0^-)-f(x_0^+)|}{3}$. By limit definition, there exists $\delta_1,\delta_2$ such that $x_0-\delta_1<x<x_0\rightarrow |f(x_0^-)-f(x)|<\varepsilon$ and $x_0<x<\delta_2+x_0\rightarrow |f(x_0^+)-f(x)|<\varepsilon$. Setting $\delta=\min(\delta_1,\delta_2)$ we obtain that
$$f\big((x_0-\delta,x_0+\delta)\big)\subset \big(f(x_0^-)-\varepsilon,f(x_0^-)+\varepsilon\big)\cup \big(f(x_0^+)-\varepsilon,f(x_0^-)+\varepsilon\big)\cup {f(x_0)}$$
But this contradicts the IVP.
We claim that a function $h:\mathbb{R}\to \mathbb{R}$ satisfies the IVP only if it maps closed intervals to intervals (in $\mathbb{R}^*$). The fact that $h$ cannot have a jump discontinuity then trivially follows.
To prove our claim, let $h([a,b])$ be a set that is not an interval (and thus it is not path connected, since the only path connected sets in $\mathbb{R}*$ are the intervals). Then there is a point $y\in[h(a),h(b)]$ that is not in $h([a,b])$, and this clearly contradicts the IVP.
To answer your other question: the IVP is not enough to ensure the existence of a primitive. Actually, it is not enough to even ensure the non boundedness of the function (see, for example, the Base 13 function).
The case in which $f=F'=k'$ is trivial. Let $F$ be a differentiable mom constant function defined on a closed interval $[a,b]$ and let $f=F'$, and let $y$ be a number in $(f(a),f(b))$ (if instead $f(b)<f(a)$ the proof is no different). Now, let $h(x):=F(x)-xy$. This function is continuous and so, by Weierstrass theorem, it has a maximum and a minimum. They can't both be on the boundary of the interval, since this would imply the fact that $F$ is constant. Thus, at least one of those point is on the interior of the interval, and by Fermat theorem in this point (let us call it $\zeta$) we have $h'(\zeta)=0\rightarrow f(\zeta)-y=0\rightarrow f(\zeta)=y$.