Let $M$ be the set of maxima of the function $k:\mathbb{R}\to \mathbb{R}$. We define the function $$L(t) = \sum_{y\in M} k(y)\delta(t-y)$$ and there by the step function $$\Gamma(t) = \int_0^t L(\tau)d\tau$$.
I'd like to know, if there is any single equation based (kind of closed form in terms of $k,\delta,k',k''$ and possibly signum function) expression without explicitly using the set $M$?
I mean is there any distributional way of writing this function?
PS : Assume $k$ is smooth!
While I still agree with my comment that there might not be a nice formula, there is something. I just don't find composing a delta function with a function very nice.
For a function $f$ with non-degenerate zeros we can naturally interpret $\delta(f(t))$ as $\sum_{a\in f^{-1}(0)}|f'(a)|^{-1}\delta(t-a)$. If $\eta$ is a smooth function, we can multiply by it to obtain $$ \eta(t)\delta(f(t)) = \sum_{a\in f^{-1}(0)}\eta(a)|f'(a)|^{-1}\delta(t-a). \tag{1} $$ In fact, it suffices that $\eta$ is smooth near zeros of $f$ (elsewhere the function vanishes). We can use this idea to write a sum over zeros without explicitly using the zero set.
If we assume that $k''\neq0$ whenever $k'=0$, we can write $$ L(t) = \sum_{y\in M}k(y)\delta(t-y) = \sum_{y,k'(y)=0}H(-k''(y))k(y)\delta(t-y) = -H(-k''(t))k(t)k''(t)\delta(k'(t)), $$ where $H$ is the Heaviside step function. I used $f=k'$ and $\eta(t)=-H(-k''(t))k(t)k''(t)$ in formula (1). I used $H$ to pick only those critical points that are local maxima, and I assumed that every critical point is a local minimum or a local maximum. Notice that the final expression contains no explicit sum.
To get $\Gamma$, you can formally integrate this expression for $L$: $$ \Gamma(t) = \int_0^tL(s)ds = -\int_0^tH(-k''(s))k(s)k''(s)\delta(k'(s))ds. $$ There might be a clever way to integrate by parts (formally) to get a nicer expression. I don't know if these formulas are of any use, but I hope they are in the desired spirit.