Let me work in homology, which is closer to the way someone like Lefschetz would have thought about it. Given a smooth complex projective variety in $X\subset \mathbb{P}^N$, we would like to understand the homology inductively. So take a general hyperplane $H\subset \mathbb{P}^N$,
then $Y=X\cap H$ is again smooth (Bertini) of smaller dimension. The Lefschetz hyperplane theorem says that $H_i(Y)\to H_i(X)$ is surjective when $i<\dim X$. So in this range,
all of the homology is effectively captured by $Y$. But when $i=\dim X$, the simple minded induction breaks down, because there will be a kernel which would be the primitive homology in the middle degree. This is the part that is genuinely new, and that needs to be understood on its own terms. This perhaps the simplest answer.
Switching to cohomology, the hard Lefschetz theorem shows that cohomology decomposes into primitive parts, so these are the essential constituents for cohomology. The hard Lefschetz theorem has an enormous number of implications for the topology of smooth projective varieties
or more generally compact Kahler manifolds. Dan's answer gives one such application. Here is another: the Betti numbers of smooth projective variety satisfy $b_0\le b_2\le b_4\ldots$
and $b_1\le b_3\le\ldots$ up to the dimension, after which they decrease. Added in response to the follow up question: I don't have the time or energy to create a new answer, so I'll comment here. The successive differences $b_3-b_1$ etc. give the dimensions of primitive cohomology.
I think that the answer to your question is no.
In order to construct a counter-example, the idea is the following: let $T$ be a Kahler current, and $E$ a $d$-closed positive current on a compact complex $3$-fold $X$ such that $T^3>0$ and $E^3<0$. Then, for $t>0$, the current $S_t:=T+tE$ is a Kahler current. Under the above conditions, you can find $t>0$ such that $S_t^3=0$. Indeed, $S_t^3=T^3+3tT^2E+3t^2TE^2+t^3E^3$ and for $t=0$, $S_0^3=T^3>0$ while for $t\to \infty$, $S_t^3<0$, so there is a $t>0$ so that $S_t^3=0$.
Now, for this $t$, denote by $\tau_t$ a smooth representative of the class of $S_t$. Then $\tau_t^3=0$. Now for instance, for $q=n=3$, the map $\tau_t^3\wedge:H^0(X,\Omega^0)\to H^3(X,\Omega^3)$ is the zero map, while $H^0(X,\Omega^0)$ and $H^3(X, \Omega ^3)$ are $1$-dimensional.
In order to fulfill these conditions, you can take $Y$ to be Hironaka's example, it is a modification of $CP^3$, so there is a map $p:Y\to CP^3$, and denote by $\omega$ the FS metric on $CP^3$
Then the $X$ mentioned above is the blow up of $Y$ at a point $x$, denote by $\pi:X\to Y$ the blow-up. If $E$ is the exceptional divisor, then $[E]$ is a positive current, and $E^3=-1<0$, and $T$ is $\pi^*p^*\omega+\varepsilon S$, where $S$ is some Kahler current on $X$ ans $\varepsilon$ is such that $T^3>0$. Then, as mentiond above, you can find $t>0$ such that $T+t[E]$ has zero self-intersection.
Best Answer
The primitive classes are the highest weight vectors.
Hard Lefschetz says that the operator $L$ (which algebraic geometers know as intersecting with a hyperplane) is the "lowering operator" $\rho(F)$ in a representation $\rho \colon \mathfrak{sl}_2(\mathbb{C})\to End (H^\ast(X;\mathbb{C}))$. The raising operator $\rho(E)$ is $\Lambda$, the restriction to the harmonic forms of the the formal adjoint of $\omega \wedge \cdot$ acting on forms. The weight operator $\rho(H)$ has $H^{n-k}(X;\mathbb{C})$ as an eigenspace (= weight space), with eigenvalue (=weight) $k$.
The usual picture of an irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$ is of a string of beads (weight spaces) with $\rho(F)$ moving you down the string and decreasing the weight by 2, and $\rho(E)$ going in the opposite direction. The highest weight is an integer $k$, the lowest weight $-k$.
From this picture, it's clear that the space of highest weight vectors in a (reducible) representation is $\ker \rho(E)$. It's also clear that, of the vectors of weight $k$, those which are highest weights are the ones in $\ker \rho(F)^{k+1}$. So the highest weight vectors in $H^{n-k}(X; \mathbb{C})$ are those in $\ker L^{k+1}$.
Of course, all this ignores the rather subtle question of how to explain in an invariant way what this $\mathfrak{sl}_2(\mathbb{C})$, or its corresponding Lie group, really is.
Added, slipping Mariano an envelope. But here's what that group is. Algebraic geometers, brace yourselves. Fix $x\in X$, and let $O_x = O(T_x X\otimes \mathbb{C})\cong O(4n,\mathbb{C})$. Then $O_x$ acts projectively on $\Lambda^\bullet (T_x X\otimes \mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol_x\cong U(n)$ also acts on complex forms at $x$, and the "Lefschetz group" $\mathcal{L}$ is the centralizer of $Hol_x$ in $O_x$. One proves that $\mathcal{L}\cong GL(\mathbb{C}\oplus \mathbb{C})$. Not only is this the right group, but its Lie algebra comes with a standard basis, coming from the splitting $T_x X \otimes\mathbb{C} = T^{1,0} \oplus T^{0,1}$. Now, $\mathcal{L}$ acts on complex forms on $X$, by parallel transporting them from $y$ to $x$, acting, and transporting back to $y$. Check next that the action commutes with $d$ and $*$, hence with the Laplacian, and so descends to harmonic forms = cohomology. Finally, check that the action of $\mathcal{L}$ exponentiates the standard action of $\mathfrak{gl}_2$ where the centre acts by scaling. (This explanation is Graeme Segal's, via Ivan Smith.)