If both mass and velocity are variable, is this the correct form for force? $v$ is velocity, $p$ is momentum, $F$ is force, $x$ is position, $t$ is time.
$$ \vec v(t) = \frac {d\vec x}{dt} $$ $$ \vec p(t) = m(t)\vec v(t) $$ $$ F = \frac {d\vec p}{dt} = \vec p'(t)= m'(t)\vec v(t) + m(t)\vec v'(t) $$ using the product rule.
with the resulting units: $$ F = \frac {kg}{s} \frac {m}{s} + kg \frac {m}{s^2} = \frac {kg *m}{s^2} + \frac {kg *m}{s^2} = \frac {2*kg*m}{s^2} $$
but when calculating $F$ without a variable mass, the units for $F$ are $F = \frac {kg*m}{s^2} = N$ which agree with what I have seen.
Therefore I think I have calculated something incorrect. Appreciate any guidance.
Your calculus is good. Your unit management is not good. You state that $$ \frac{kg \cdot m}{s^2} + \frac {kg \cdot m}{s^2} = \frac {2\cdot kg\cdot m}{s^2} $$ but this is wrong.
As an example, think about adding two lengths together $\ell_1$ and $\ell_2$. If they both have units of meters then the units on the resulting sum are not "2 meters" but just meters. In other words
$$ (\ell_1 \text{ meters }) + (\ell_2 \text{ meters }) = (\ell_1 + \ell_2) \text{ meters } \neq (\ell_1 + \ell_2)\: \: 2\text{ meters. } $$
Likewise the correct way to add units of force is to say
$$ \frac {kg \cdot m}{s^2} + \frac {kg \cdot m}{s^2} = \frac {kg\cdot m}{s^2} \text{ .}$$