meaning of dual space of a tangent space?
Here is an example. If $M$ is a manifold and $f : M \to \mathbb{R}$ a scalar function, then for a point $p \in M$ the derivative (or gradient) $df_p : T_p(M) \to \mathbb{R}$ is naturally a cotangent vector (an element of the dual of the tangent space). It measures how much $f$ changes in the direction of a given tangent vector. For example, if $M$ is a room and $f$ is the temperature of a point in the room, then $df_p$ measures how quickly the temperature is changing in a particular direction.
There are a few ways to look at cotangent vectors (the things dual to tangent vectors). Depending on what position you're trying to take, some are more natural than others.
At the most basic level, if you have a manifold of dimension $n$, and $p\in M$, then $T_pM \cong T^*_p M \cong \mathbb{R}^n$. However, this isomorphism isn't telling you anything about where all of this comes from.
A slightly better answer comes by noticing that tangent vectors act on real valued functions. Let $v$ be a tangent vector at $p$, and let $f$ be a function. First, we note that $v\mapsto v(f)$ is a linear function, so $f$ determines an element of the $T_p^*M$. Also $v(f)=v(f+c)$, and so if we are trying to understand this action, we might as well look at functions that satisfy $f(p)=0$. Now, let $f,g$ be functions such that $f(p)=g(p)=0$. Then by the product rule $v(fg)=v(f)g(p)+f(p)v(g)=0$.
We can rephrase all this as follows: If we let $\mathfrak{m}_p \subset C^{\infty}(M)$ be the ideal of functions vanishing at $p$, and let $df_p\in T^*_pM$ be such that $df_p(v)=v(f)$. Then the constant functions, and $\mathfrak{m}_p^2$ are both in the kernel of the map $d(-)_p$.
However, because tangent vectors only care what a function is doing locally, we don't really want to be looking at the action of $v$ on functions, but rather on germs of functions, which codify the idea of having a function defined only locally. Changing everything above to be in terms of germs (so that $\mathfrak{m}_p$ is the collection of germs of functions at $p$ that vanish at $p$), we have a natural map $\mathfrak{m}_p/\mathfrak{m}_p^2 \to T_p^*M$. By using the fundamental theorem of calculus, we can show that this is an isomorphism.
So cotangent vectors are equivalence classes of (germs of) functions, where two functions are equivalent if they differ by either a constant function or a function that vanishes at $p$ with multiplicity at least $2$. In this sense, they are not operators, but they do act on tangent vectors, by allowing the tangent vectors to act on them.
A slightly more advanced point of view is to put all the tangent spaces together, and all the cotangent spaces together. If you take a tangent vector at every point (and you do this in such a way that the vector you choose varies smoothly), you get the notion of a vector field. Vector fields act on functions to give functions. Similarly, if you pick a cotangent vector at every point (in such a way that the vector varies smoothly), you get the notion of a differential ($1$-)form. Differential forms act on vector fields to give functions. In some sense, this action sets up $TM$ and $T^*M$ as duals to each other, but duals with respect to the ring of functions $C^{\infty}(M)$, not vector space duals. This does paint differential forms as being operators of a kind.