As already noted by @Chet Miller, $\dot Q=\dot q A$. Therefore the first equation can be written as
$$R=\frac{\Delta T}{\dot Q}=\frac{\Delta T}{\dot q A}=\frac{L}{kA}$$
making it identical to the second equation when multiplying each side by $A$.
But it strikes me as odd, since in electric circuits the units are
always Ohms.
$R$ in these equations is thermal resistance with units of Kelvins per watt ($K/W$). The electrical analog for steady heat transfer through a wall is Ohms law which gives the relationship between current ($I$) voltage ($V$) and resistance ($R$), where $R$ is electrical resistance with units of Ohms ($\Omega$).
With regard to the electrical analogs for the other parameters:
$k$ is thermal conductivity with units watts per meter Kelvin ($W/(m\cdot K$)). The electrical analog is electrical conductivity with units ohm meters ($\Omega\cdot m$).
$\Delta T$ is the temperature difference between the two walls in degrees Kelvin ($K$). The electrical analogy is potential difference if volts ($V$), or Joules per Coulomb ($J/C$)
$\dot Q$ is heat transfer rate in watts, or Joules per second ($J/s$). The electrical analogy is current in amperes ($A$) or Coulombs per second ($C/s$)
$\dot q$ is heat flux in watts per square meter ($W/m^2$). The electrical analog is current density in amperes per square meter ($A/m^2$).
Hope this helps.