OK, so I am not an astrophysicist, or anywhere near one, but the following is cobbled together from memory and some looking things up on Wikipedia (links below).
First of all if you naively assume that a star is (a) spherical and (b) a black body then you have this important relation:
$$L = 4\pi R^2\sigma T^4\tag{1}$$
Where $R$ is radius, and $T$ is temperature, $\sigma$ is the Stefan-Boltzmann constant and $L$ is the total luminosity (power output). The $4\pi R^2$ is the formula for the surface area of a sphere, of course.
Secondly there is a horrible thing called the mass-luminosity relation, which is an observed relation for main-sequence stars. It can (I am sure) be derived from models of how stars work, but here it is in a form which is suitable for 'engineering use':
$$\tag{2}
\frac{L}{L_\odot} \approx \begin{cases}
0.23\left(\frac{M}{M_\odot}\right)^{2.3}&M < 0.43M_\odot\\
\left(\frac{M}{M_\odot}\right)^{4}&0.43M_\odot \le M < 2M_\odot\\
1.4\left(\frac{M}{M_\odot}\right)^{3.5}&2M_\odot \le M < 55M_\odot\\
32000\frac{M}{M_\odot}&M \ge 55M_\odot
\end{cases}
$$
Where $L$ is total luminosity, $M$ is mass, and $L_\odot, M_\odot$ are the luminosity and mass of the Sun respectively.
This comes directly from this Wikipedia page.
Finally there is another 'good enough for engineering' relation for main-sequence stars:
$$R \propto M^{0.78}\tag{3}$$
So, the first things to start from are knowing $M_\odot$, $L_\odot$ and $R_\odot$, which you can look up. Given these you can work out the proportionality constant in (3) easily enough.
Then given the mass of the star you can use (2) to work out its luminosity. And finally you can use the $R$ you get from (3) together with (1) to work out its surface temperature.
There is another important thing to bear in mind: the spectrum of the light emitted by a black body depends on the surface temperature, and Wein's displacement law says that the wavelength for the peak intensity is
$$\lambda_p = \frac{b}{T}$$
where $b \approx 2.9\times 10^{-3}\,\mathrm{m\,K}$ (see Wikipedia again).
If you are planning on carbon-based biological life in your imagined stellar system, you don't want the wavelength of the light emitted by the star to be too short, and that means you don't want the temperature too high. If the temperature is too high then you'll get a lot of ionizing radiation and this will pull complex organic molecules to bits. So this places some kind of upper bound on $T$ (picking something similar to the Sun's temperature is probably a safe idea).
So the place to start is probably with the temperature of the star,and the assumption that it is main-sequence. If you want carbon-based life you can assume that $T$ in some suitable range: you don't want it too low because you want some UV, you don't want it too high because you don't want too much UV or x-rays or whatever. And stars spend most of their time on the main sequence, so that's where you have time for life to evolve.
If you additionally assume that $0.43M_\odot \le M < 2M_\odot$ (just to make using (2) easier), then, from (1):
$$\frac{L}{R^2} = 4\pi\sigma T^4$$
And then we can use the relation from (2): with the assumption that $0.43M_\odot \le M < 2M_\odot$, we have that $L \approx (M/M_\odot)^4 L_\odot$, and we can substitute this expression for $L$ into the previous equation to get this:
$$\frac{M^4 L_\odot}{M_\odot^4 R^2} \approx 4\pi\sigma T^4$$
or, moving constant terms to the RHS:
$$\frac{M^4}{R^2} \approx \frac{4\pi\sigma M_\odot^4}{L_\odot} T^4$$
Now, writing (3) as $R \approx K M^{0.78}$, where we need to work out $K$ by looking at the Sun later, we have $R^2 \approx K^2 M^{1.56}$, so
$$\frac{M^{2.44}}{K^2} \approx \frac{4\pi\sigma M_\odot^4}{L_\odot} T^4$$
And finally this gives us an expression for $M$ in terms of $T$:
$$M \approx \left(\frac{4\pi\sigma M_\odot^4 K^2}{L_\odot} T^4\right)^{0.41}$$
So this is an expression for $M$ in terms of $T$. So, given $T$ you can work out $M$, and then given $M$ you can work out $R$ and thus $L$.
There may be mistaked in the above maths, and the expression is clearly dimensionally a bit mad, probably because it's an approximation to something much more complicated.
Finally some more pointers (all of this is to Wikipedia, sorry):
Obviously there are both better and more in-depth places to find this information, but Wikipedia is not hopeless in my experience.
