If I interpret the request a bit differently, I would say that the Steenrod operations in the cohomology of a spectrum tell you about the attachments of the cells. If $Sq^1 x = y$, then a cell dual to $y$ is attached by a map of degree 2 mod 4 to a cell dual to $x$. Similarly, $Sq^2 x = y$ tells us the attaching map is $\eta$, $Sq^4$ detects $\nu$ and $Sq^8$ detects $\sigma$. This doesn't go very far, but may help with the need to 'get a real grip on what they're doing'.

Next, let's assume you're really interested in homotopy, not just (co)homology. A class dual to a homology class in the image of the Hurewicz homomorphism must be indecomposable under the action of the Steenrod algebra, by naturality w.r.t. the map $S^n \longrightarrow X$. This limits the homotopy of $X$ which can be detected by the homomorpism $\pi_* X \longrightarrow H_* X$: the homomorphism $H^* X \longrightarrow H^* S^n$ can only map indecomposables non-trivially, since all classes in degrees below $n$ must go to $0$.

Then there are the relations. The fact that $Sq^n$ is decomposable when n is not a power of two tells us that if $y = Sq^n x$, there must be other classes between $x$ and $y$. EG, $Sq^3 x = y \neq 0$ tells us that $Sq^2 x \neq 0$ also, since $Sq^3 = Sq^1 Sq^2$. So our spectrum can't have just two cells, dual to $x$ and $y$, but must have a three cell subquotient with top cell attached by 2 (mod 4) to a cell attached by $\eta$ to the bottom cell.

Or, if $Sq^2 Sq^2 x = y \neq 0$ then we must also have nonzero classes $Sq^1 x$ and $Sq^2 Sq^1 x$, since $Sq^2 Sq^2 = Sq^1 Sq^2 Sq^1$, and vice versa, if $Sq^1 Sq^2 Sq^1 x = y \neq 0$ then $Sq^2 x \neq 0$ as well. This leads to an easy proof that the mod 2 Moore spectrum $M$ isn't a ring spectrum, since $2 \pi_0M = 0$ but $\pi_2 M = Z/4$, by looking at the obstruction to attaching the top cell of a putative spectrum with nonzero cohomology spanned by $x$, $Sq^1 x$, $Sq^2 Sq^1 x$, and $Sq^1 Sq^2 Sq^1 x$. More, the fact that you can only add such a top cell if you also have a class $Sq^2 x$ so that the top cell can be attached by the sum of $Sq^1$ on $Sq^2 Sq^1 x$ and $Sq^2$ on $Sq^2 x$ shows that $\eta^2$ (corresponding to the path $Sq^2 Sq^2$ from bottom to top, must lie in the Toda bracket $\langle 2, \eta, 2\rangle$, corresponding to the path $Sq^1$, $Sq^2$, $Sq^1$ from bottom to top.

Similarly, $y = Sq^1 Sq^2 x$ tells us that homotopy supported on a cell dual to $x$ can be acted on by $\text{v}_1$ to get $y$, literally if we have a $ku$-module and multiply by $\text{v}_1 \in ku_2$, or as the Toda bracket $\langle 2, \eta, -\rangle$ more generally. The key fact here is that $\text{v}_1 \in ku_2$ is in $\langle 2, \eta, 1_{ku} \rangle$, where $1_{ku} : S \longrightarrow ku$ is the unit.

Likewise, $Sq^2 Sq^1 Sq^2 x = y$ corresponds to multiplication by the generator of $ko_4$, literally for $ko$-modules, or as a bracket $\langle \eta, 2, \eta, - \rangle$ more generally. Here you have to be in a situation where $2 \nu = 0$ to form the bracket, since $\langle \eta, 2, \eta \rangle = \{ 2\nu, 6 \nu\}$. This hints that the role of $\nu$ is non-trivial in real K-theory, despite going to $0$ under the homomorphism $\pi_* S \longrightarrow \pi_* ko$ and despite the cohomology of $ko$ being induced up from the subalgebra $A(1)$ generated by $Sq^1$ and $Sq^2$. The Adem relation $Sq^2 Sq^1 Sq^2 = Sq^1 Sq^4 + Sq^4 Sq^1$ shows that $Sq^4$ must act nontrivially if $Sq^2 Sq^1 Sq^2$ does. Also, the fact that $A(1)//A(0)$ is spanned by $1$, $Sq^2$, $Sq^1 Sq^2$, and $Sq^2Sq^1Sq^2$ tells us (with a bit more work) that we can build $HZ$ as a four cell $ko$-module.

A good way to organize all this information is the Adams spectral sequence, which tells you that the mod $p$ cohomology of $X$ gives a decent first approximation, $\text{Ext}_{A}(H^*X,F_p)$, to the homotopy of the $p$-completion of $X$.