Here it will be shown how conditions on the $f_n$'s or on $f$ can be modified to make the sequence $(x_n)$ convergent.

First of all, the condition that the $f_n$'s be twice differentiable is of no help; so, this condition will be dropped.

Suppose that for each $n$ there exists some $x_n\in\mathbb{R}$ such that:

$f_n$ is semi-strongly convex on $(-\infty,x_n)$ in the sense that $f_n$ is convex on $(-\infty,x_n)$ and $f_n(x)/|x|\to\infty$ as $x\to-\infty$;

$f_n$ is semi-strongly concave on $(x_n,\infty)$ in the sense that $f_n$ is concave on $(x_n,\infty)$ and $f_n(x)/x\to-\infty$ as $x\to\infty$.

Note that, if $f_n$ is strongly convex on $(-\infty,x_n)$, then $f_n$ is semi-strongly convex on $(-\infty,x_n)$; similarly, if $-f_n$ is strongly convex on $(x_n,\infty)$, then $f_n$ is semi-strongly concave on $(x_n,\infty)$.

Suppose also that $(f_n)$ uniformly converges to a function $f$, which
is not affine on any nonempty interval.

Then the sequence $(x_n)$ is convergent.

Indeed, suppose first that the sequence $(x_n)$ is unbounded. Then, using a left-right symmetry and passing to a subsequence, without loss of generality (wlog) assume that $x_n\to\infty$. It follows that the limit function $f$ is convex on $\mathbb R$. So, $f(x)\ge a+bx$ for some real $a,b$ and all real $x$. Therefore and in view of semi-strongly concavity of $f_n$ on $(x_n,\infty)$, for each $n$ we have $f_n(x)-f(x)\to-\infty$ as $x\to\infty$, which contradicts the uniform convergence of $(f_n)$ to $f$.

So, the sequence $(x_n)$ is bounded. Take any subsequence $(x_{n_k})$ of the sequence $(x_n)$ converging to a limit $x_*$. Then $f$ is convex on $(-\infty,x_*)$ and concave on $(x_*,\infty)$. Since $f$ is not affine on any nonempty interval, the point $x_*$ is uniquely determined by $f$. So, any converging subsequence of the sequence $(x_n)$ converges to the same limit $x_*$. Since the sequence $(x_n)$ is bounded, we conclude that $x_n\to x_*$, as claimed.

This reasoning shows that the following is true as well:

Suppose that for each $n$ there exists some $x_n\in\mathbb{R}$ such that:

$f_n$ is convex on $(-\infty,x_n)$;

$f_n$ is concave on $(x_n,\infty)$.

Suppose also that $(f_n)$ converges pointwise (not necessarily uniformly) to a function $f$, which
is not affine on any nonempty interval, not convex on $\mathbb R$, and not concave on $\mathbb R$.

Then the sequence $(x_n)$ is convergent.

**Details on the latter assertion, requested by Gaetano:** Suppose first that the sequence $(x_n)$ is unbounded. Then, using a left-right symmetry and passing to a subsequence, wlog assume that $x_n\to\infty$. It follows that the limit function $f$ is convex on $\mathbb R$, which contradicts the assumption that $f$ is not convex on $\mathbb R$.

So, the sequence $(x_n)$ is bounded. Take any subsequence $(x_{n_k})$ of the sequence $(x_n)$ converging to a limit $x_*$. Then $f$ is convex on $(-\infty,x_*)$ and concave on $(x_*,\infty)$. Since $f$ is not affine on any nonempty interval, the point $x_*$ is uniquely determined by $f$. So, any converging subsequence of the sequence $(x_n)$ converges to the same limit $x_*$. Since the sequence $(x_n)$ is bounded, we conclude that $x_n\to x_*$, as claimed.