I will give an answer for the simpler case of the free Wess-Zumino (WZ) model and then you can try to do the same for your SUSY Yang-Mills theory. The free WZ model has the following Lagrangian

$$ \mathcal{L} = \partial^{\mu} \phi \partial_{\mu} \phi^* + i\psi^{\dagger} \bar{\sigma} \partial_{\mu}\psi $$

Let us consider the infinitesimal SUSY transformations $\phi \to \phi + \delta \phi$ and $\psi \to \psi + \delta \psi$. Now, SUSY is the symmetry that says that under a small variation the bosons go to fermions and the fermions go to bosons. Therefore it is only natural that the change of the boson is a fermion. We set $\delta \phi = \epsilon^{\alpha}\psi_{\alpha} = \epsilon \psi$, where $\alpha$ are sponsorial indices and as you see this makes sense in terms of dimensions as well if we require that $\epsilon$ is an infinitesimal Grassman variable with dimension -1/2. It is not hard to show that $\delta \phi^* = \epsilon_{\dot{\alpha}}^{\dagger} \psi^{\dagger \,\dot{\alpha}} = \epsilon^{\dagger} \psi^{\dagger}$. Then one can ask, what is a small SUSY variation of the Lagrangian? The answer is

$$ \delta \mathcal{L}= \epsilon \partial^{\mu} \psi \partial_{\mu} \phi^{*} + \epsilon^{\dagger} \partial^{\mu}\psi^{\dagger} \partial_{\mu} \phi $$

Now we want $\delta \mathcal{L}_s$ to be zero. But since the the first term involves two derivatives and the first one only one it means we must modify the SUSY transformation for the fermion field. To do it properly we need the following transformation

$$ \delta \psi_{\alpha} = -i(\sigma^{\nu} \epsilon^{\dagger})_{\alpha} \partial_{\nu} \phi\,\,\, $$

and

$$ \delta \psi_{\dot{\alpha}}^{\dagger} = i(\epsilon \sigma^{\nu})_{\dot{\alpha}} \partial_{\nu}\phi^{*} $$

Then the change in $\mathcal{L}$ is given by

$$ \delta \mathcal{L} = -\epsilon \sigma^{\nu} \partial_{\nu} \phi^* \bar{\sigma}^{\mu} \partial_{\mu} \psi+ \psi^{\dagger} \bar{\sigma}^{\mu} \sigma^{\nu} \epsilon^{\dagger} \partial_{\mu} \partial_{\nu} \phi$$.

Then, after performing a few integrations by parts, you can see that all terms cancel out, giving $\delta \mathcal{L}=0$.

The SUSY transformations are not derived by the SUSY algebra, rather than the opposite. Now, since we know what the SUSY transformations are we ask ourselves what happens with the commutator of two SUSY transformations $\delta_1$ and $\delta_2$? Then we consider

$$ [\delta_1, \delta_2] = -i(\epsilon_1\sigma^{\mu} \epsilon_2^{\dagger} - \epsilon_2 \sigma^{\mu}\epsilon_1^{\dagger} ) \partial_{\mu} $$

which as you can see it really resembles the equation

$$ [Q_{\alpha}, {Q}_{\beta}^{\dagger}] \backsim \sigma^{\mu}P_{\mu} \delta_{\alpha \beta}$$.

Then we can do the same, find the commutator of two SUSY transformations for the fermion. If we do so, and after using the Fiery identity we find a similar expression, i.e. a proportionality to the momentum operator. This is why we say that the commutator of two SUSY transformations results into a boost. Additionally it shows that the SUSY algebra closes on-shell.

Now, all you have to do is to consider instead of a scalar field $\phi$, a vector $A_{\mu}$. The idea is the same, except that for in both cases when you want to go off-shell you need to consider additional auxiliary field in order the SUSY algebra to be closed. The details can be found in many books and references like Martin's lecture notes, Wess and Bagger's book, John Terning's book, Van Proyen's Supergravity book and so on. But in any case, the generalisation to SUSY Yang-Mills is straight forward.

P.S. By the way, under SUSY the gauge field transforms under representations of the corresponding super-Lie algebra, not just the SUSY algebra. Each member of the multiplet must transform in the same gauge representation, but I am not really sure if you do consider only representations for SUSY neglecting the gauge part.