With the aid of machine computations, you can readily determine the Adams differentials up to $t-s=30$ using the multiplicative structure, the relation between Steenrod operations in $\text{Ext}_A$ and the $E_\infty$ (or $H_\infty$) ring spectrum structure, a comparison with the mapping cone $C_\sigma = S \cup_{\sigma} e^8$ and a comparison with the image-of-$J$ spectrum $j = \text{hofib}(\psi^3-1 : ko \to bspin)$.

In this limited range this sidesteps the use of Toda brackets and the solution of the Adams conjecture, which were frequently used by Barratt, Mahowald and Tangora for their calculation in the range $t-s \le 45$.  This is not an argument for not learning about those methods and results!  Nonetheless, this is how I recently went through the calculation in class:

In the range $t-s \le 30$ the Adams $E_2$-term is generated as an algebra by the $h_i$ for $0 \le i \le 4$ in Adams filtration $s=1$, $c_0$ and $c_1$ in filtration $s=3$, $d_0$, $e_0$, $f_0$ and $g = g_1$ in filtration $s=4$, $Ph_1$ and $Ph_2$ with $s=5$, $r$ with $s=6$, $Pc_0$, $i$, $j$ and $k$ with $s=7$, $Pd_0$ and $Pe_0$ with $s=8$, $P^2h_1$ and $P^2h_2$ with $s=9$, $P^2c_0$ with $s=11$, $P^2d_0$ with $s=12$, and $P^3h_1$ and $P^3h_2$ with $s=13$.  This can easily be checked with Bruner's ext-program. The notation is derived from Peter May's thesis, as presented in Martin Tangora's 1970 Math. Z. paper.  Only $f_0$ is ambiguously defined modulo decomposables.

Many differentials can now be determined by the relation between the $E_\infty$ structure on $S$ and the Steenrod operations on $\text{Ext}_A$.  This is due to D.S. Kahn, J. Milgram, J. Makinen and R.R. Bruner, in increasing generality. See Bruner's chapter in the $H_\infty$ book of Bruner, May, McClure and Steinberger for further details.

The $E_\infty$ structure applied to $\sigma : S^7 \to S$ gives a map $\Sigma^7 RP_7^\infty = D_2(S^7) \to S$ that extends $\sigma^2$, and gives you the relations $2\sigma^2 = 0$ (which forces the differential $d_2(h_4) = h_0h_3^2$) and $\eta\sigma^2 = 0$.

The $E_\infty$ structure applied to $2\sigma : S^7 \to S$, together with the Steenrod operations on $\text{Ext}_A$, tells you that $h_1 h_4$ is a permanent
cycle.

The $E_\infty$ structure applied to $\eta\sigma : S^8 \to S$ tells you that $h_2 h_4$ is a permanent cycle, if you accept a Cartan formula for power operations.  This deduction can be replaced by the arguments below.

The $E_\infty$ structure applied to $\epsilon : S^8 \to S$ gives the differential $d_2(f_0) = h_0^2 e_0$, and tells you that $c_1$ is a permanent cycle.  This relies on the algebraic facts that $Sq^2(c_0) = h_0 e_0$,
$Sq^1(c_0) = f_0$ and $Sq^0(c_0) = c_1$.  (Bruner's original paper uses a
different indexing system for the $Sq^i$ than I use here.)

Using $h_0$- and $h_1$-linearity of $d_2$, you can deduce $d_2(h_0 f_0) =
h_0^3 e_0$ and $d_2(e_0) = h_1^2 d_0$.

Multiplying by $d_0$, you get $d_2(d_0f_0) = h_0^2 d_0 e_0$, and using
$h_2$-, $h_0$- and $h_1$-linearity of $d_2$ you find $d_2(k) = h_0 d_0^2$, $d_2(j) = h_0 Pe_0$, $d_2(Pe_0) = h_1^2 i$ and $d_2(i) = h_0 Pd_0$.  Many more $d_2$-differentials for higher $t-s$ follow in this way.

Together with the Adams differential $d_2(h_5) = h_0 h_4^2$, these are the only $d_2$'s affecting $t-s \le 30$. This is clear from the multiplicative structure (and the fact that $d_2 \circ d_2 = 0$).

To get at the $d_3$-differentials, you can use naturality along $i : S \to C_\sigma$.  There is a class $\beta$ in bidegree $(t-s,s) = (15,3)$ of the $E_2$-term of the Adams spectral sequence for $C_\sigma$, such that $h_2 \beta
= i_*(f_0)$.  (This notation is compatible with the $tmf$-notation used by A. Henriques and T. Bauer.)  By naturality $d_2(i_*(f_0)) = h_0^2 i_*(e_0)$, and by $h_2$-linearity $d_2(\beta) = h_0 i_*(d_0)$ in the Adams spectral sequence for $C_\sigma$.  Thus $d_2(h_0 \beta) = h_0^2 i_*(d_0)$.  It follows that $\pi_{14}(C_\sigma)$ (implicitly $2$-completed) has order dividing $2^2=4$.

Looking at the long exact sequence
$$
\pi_7(S) \overset{\sigma}\longrightarrow \pi_{14}(S) \to \pi_{14}(C_\sigma) \to \pi_6(S) \to \pi_{13}(S)
$$
you can deduce that $\pi_{14}(S)$ has order exactly $4$, so that $d_0$ and
$h_3^2$ survive to $E_\infty$ and detect $\kappa$ and $\sigma^2$, respectively. This also implies that $h_0 d_0$ must be a boundary, and $d_3(h_0 h_4) = h_0 d_0$ is the only possibility, which implies $d_3(h_0^2 h_4) = h_0^2 d_0$.

You can also deduce that $i_*(h_4)$ is a permanent cycle for $C_\sigma$,
and this implies that $d_3(h_2h_4) = 0$ for $S$.

The final $d_3$-differential in this range, $d_3(r) = h_1 d_0^2$, can be obtained from the $E_\infty$ structure applied to $\kappa : S^{14} \to S$, using $Sq^2(d_0) = r$.  The next possible $d_3$-differentials start in
$t-s=31$.

The only possibly nonzero $d_4$-differential is that on $h_4 c_0$, which might hit $Pd_0$, but $h_1 h_4 c_0$ is a permanent cycle and $h_1 Pd_0$ is nonzero, so this is excluded by $h_1$-linearity of $d_4$.

The only possibly nonzero later differentials are those on $h_2 h_4$ and $g$, which might conceivably hit $h_1 Pc_0$ or an $h_0$-power times $Ph_2$, respectively. To eliminate this possibility, one can use naturality with respect to the map $e : S \to j$.

The cohomology of $j$ is the unique nonsplit extension of the $A$-modules given by the cokernel and desuspended kernel of
$$
(\psi^3-1)^* : \Sigma^4 A/A\{Sq^1, Sq^2Sq^3\} = H^*(bspin) \to H^*(ko) = A/A\{Sq^1, Sq^2\} \,,
$$
sending the generator $\Sigma^4 1$ to $Sq^4$.  Its Adams $E_2$-term can be calculated in this range by Bruner's programs, and in the range $t-s \le 24$ it is easy to see that the only differential pattern that is compatible with the known abutment, i.e., $\pi_*(j)$ completed at $2$, is one with $E_3 = E_\infty$.

The map of $E_3$-terms induced by $e : S \to j$ is surjective for $t-s \le 24$, except for $t-s=15$ (the homomorphism $\pi_{15}(e)$ involves a shift of Adams filtration).  In particular, the imagined targets for differentials on $h_2 h_4$ and $g$ in the Adams spectral sequence for $S$ map to survivors in the Adams spectral sequence for $j$, hence cannot be hit by differentials for $S$, after all.