If a force $F_0$ acts on a movable particle for a short time, we get (setting integrations constants to 0)
$\ddot{x} = \frac{F_0}{m}\delta(t-t_0) \\ \dot{x} = \frac{F_0}{m}H(t-t_0) $
So does the Heaviside function implicitly have the same unit as its argument or is this just wrong?
Let us first analyze the dimensions of the $\delta$-function. From the definition of $\delta$-function
$$\int d\xi\,\delta(\xi) = 1.$$
For the dimensions of the LHS and RHS to be equal, there can only be two possibilities: (i) $ξ$ is dimensionless, in which case $δ(ξ)$ must also be dimensionless, or (ii) $ξ$ has dimensions, in which case $δ(ξ)$ must have dimensions of $ξ^{-1}$.
In your example, the argument of the delta function has dimensions of time, thus it would seem that your Eq. (i) has incorrect dimensions. One way to get around this issue is to assume that there is an imaginary "$1$" inside the delta function to fix the units, i.e., assume that
$$ δ(t-t_0) = \delta\left(\frac{t-t_0}{1 \text{ time unit}}\right).$$
If we use this definition, then $\delta(t)$ is dimensionless and Eq. (i) makes sense. Now, the Heaviside step function is defined by $$H(x) = \int_{-\infty}^x \delta(\xi)\,d\xi.$$ For your example,
$$ H(t) = ∫_{-∞}^{t} ds\, \delta\left(\frac{s}{1 \text{ time unit}}\right) = (1 \text{ time unit})\times \int_{-\infty}^{t/(1 \text{ time unit})} ds'\,\delta(s'),$$
which has dimensions of time, as the integral is completely dimensionless. Thus, Eq. (ii) also makes sense once we assume the existence of an imaginary "$1$" with the correct units.
You could also assume that $F_0$ and $m$ have nonstandard dimensions so that $F_0/m$ has dimensions of $\text{time}^{-1}$ (e.g., assume that $F_0$ is an impulse instead of a force). In this case, $\delta(t)$ would have dimensions of $\text{time}^{-1}$ and the Heaviside step function $H(t)$ would be dimensionless.