Your expression of the final state and your computation of the probability of measuring $0$ are both correct. We thus have to find the expression of $\left|\psi_f\right\rangle$ once the first qubit has been measured as $|0\rangle$.
Mathematically, up to a normalisation factor, the expression of $\left|\psi_f\right\rangle$ is equal to:
$$\left(|0\rangle\!\langle0|\otimes I_2\right)|\psi\rangle$$
where $I_2$ is the $2\times2$ identity matrix and $|\psi\rangle$ is the state just before the measurement.
When I started to learn Quantum Computing, I found it quite difficult to deal with the mathematical formalism. At the beginning of your journey, you will essentially deal with pure states and measurements performed in the basis in which you've expressed your state. The following method will allow you to find the final expression of your state once a subsystem has been measured.
We start from:
$$|\psi\rangle=\frac12\left[(a+b)|00\rangle+(a+b)|01\rangle+(a-b)|10\rangle+(b-a)|11\rangle\right]$$
Thus, $|\psi\rangle$ is a superposition between four terms. What happens during the measurement of the first qubit is that the state will collapse. This collapse will only keep the terms that are consistent with the measurement. In this case, it will only keep the terms in which the first qubit is equal to $0$:
$$|0\rangle\left|\psi_f\right\rangle\propto\frac12\left[(a+b)|00\rangle+(a+b)|01\rangle\right]$$
Note that I didn't write an equality symbol, but a proportionality one. Indeed, the state we've written is not a valid quantum state since its norm isn't one. Now that the first qubit isn't entangled anymore with the rest of the system, let us put it out of the expression:
$$\left|\psi_f\right\rangle\propto\frac12\left[(a+b)|0\rangle+(a+b)|1\rangle\right]$$
Now, we simply have to compute the norm of this "pseudo-state" so that we can normalise it:
$$\left\|\frac12\left[(a+b)|0\rangle+(a+b)|1\rangle\right]\right\|=\frac{a+b}{\sqrt{2}}$$
Thus, the final expression for $\left|\psi_f\right\rangle$ is:
$$\left|\psi_f\right\rangle=\frac{\frac12\left[(a+b)|0\rangle+(a+b)|1\rangle\right]}{\frac{a+b}{\sqrt{2}}}=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$
As a side note, we could have been a little bit faster by noting that $\left|\psi_f\right\rangle$ is proportional to $\frac{a+b}{2}(|0\rangle+|1\rangle)$, which means that $|0\rangle$ and $|1\rangle$ share the same amplitude. You can always remove the global normalisation factor (in this case, $\frac{a+b}{2}$) because you're going to normalise the state anyway, but it is important to keep the relative amplitudes intact.
For instance, if you start with the state:
$$\frac{|00\rangle+2|01\rangle+|11\rangle}{\sqrt{6}}$$
If you measure $|0\rangle$ on the first qubit, the state of the second qubit is proportional to $|0\rangle+2|1\rangle$, which means that it is $\frac{1}{\sqrt{5}}|0\rangle+\frac{2}{\sqrt{5}}|1\rangle$.