The polynomial $Q(y):=\frac43y-\frac13 {y^3}$ is increasing on the interval $[-1,1]$; it has fixed points $0$ and $\pm1$, and $\text{sgn }( Q(y)-y )=\text{sgn}y$. Thus the iterates of $Q$ starting from any $y\in [-1,1]\setminus\{0\}$ converge monotonically to $\text{sgn } y$ (in fact with exponential rate given by $Q'(\pm1)=\frac13$, and uniformly away from $0$
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Assuming w.l.o.g $a_i=(-1)^i$, your step function can be written $f(x):= \text{sgn} P(x)$ with $P(x):= \prod_{i=1}^m(t_i-x)$, for any $x\in X$. If we choose any $M\ge \|P(x)\|_{\infty,X}$ the polynomial sequence of iterates $Q^{n}\big( {P(x)}/ M\big)$ converges to $f$ on $X$; in fact increasing/decreasing between consecutive nodes, and uniformly on compacts set of $X$.
Claim: The iterated sum $f_k(t_1,\ldots,t_k)$ counts the number of elements the interval $[\emptyset,\lambda]$ of Young's lattice, where $\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_k)$ is the partition determined by $\lambda_{k-i+1} = t_1 + \cdots + t_i$. Equivalently, the function $f_k$ counts the number of subdiagrams of $\lambda$.
For an arbitrary partition $\lambda$, we have
$$|[\emptyset,\lambda]| = \text{det} \left[\binom{\lambda_i + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$
which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of Stanley's Enumerative Combinatorics, volume 1, 2nd edition provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindström-Gessel-Viennot, see Ciucu -
A short conceptual proof of Narayana's path-counting formula.
If the claim is true, MacMahon's result implies
$$\sum_{j_1=0}^{t_1}\sum_{j_2=0}^{t_2+j_1}\cdots\sum_{j_k=0}^{t_k+j_{k-1}} = \text{det} \left[\binom{t_1 + \cdots + t_{k - i + 1} + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$
which implies $f_k(t_1,\ldots,t_k)$ is a polynomial in $t_1,\ldots,t_k$.
Note that $f_k(t_1,\ldots,t_k)$ counts the number of $(j_1,\ldots,j_k)$ such that $0 \leq j_1 \leq t_1$ and $0 \leq j_{i+1} \leq j_i + t_{i+1}$ for $i \geq 1$. To establish the claim, it suffices to find a bijection between the set of $\mu \subseteq \lambda$ and the set of tuples satisfying the above constraints.
Sketch: Map $\mu \subseteq \lambda$ to $(j_1,\ldots,j_k)$, where $j_i = \lambda_{k-i+1} - \mu_{k-i+1}$. The visual interpretation is that each $j_i$ measures the distance between the walls of the $i$-th row from the bottom of the Young diagrams (English convention) for $\mu$ and $\lambda$. The $t_i$ specify how many boxes are added to the diagram for $\lambda$ in moving from the $(i-1)$-st row from the bottom to the $i$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $t_i$. For a more direct definition chase, note that $\lambda_{k-i} - \lambda_{k-i+1} = t_{i+1}$. Since $\mu$ is a partition, we have $\mu_{k-i+1} - \mu_{k-i} \leq 0$. Combining the definitions and inequalities gives $j_{i+1} \leq j_i + t_{i+1}$.
Best Answer
If you have two polynomials $f(x)=a_0+a_1x+\cdots a_mx^m$ and $g(x)=b_0+b_1x+\cdots+b_nx^n$, such that the roots of $f$ are all real, and the roots of $g$ are all real and of the same sign, then the Hadamard product $$f\circ g(x)=a_0b_0+a_1b_1x+a_2b_2x^2+\cdots$$ has all roots real. This was proven originally in
One can make stronger statements, such as the result by Schur that says that $\sum k!a_kb_k x^k$ will only have real roots, under the same conditions. Schur's theorem combined with the fact that $\{1/k!\}_{k\geq 0}$ is a Polya frequency sequence, implies Malo's theorem.
I'm not sure what the best reference to learn the theory of real rooted polynomials, and the associated operations that preserve real rootedness is. One textbook I know that discusses some of these classical results is Marden's "Geometry of Polynomials".