We know that we can define the de-Rham cohomology because $d^2=0$. However, we also have the property $\iota_X \iota_X=0$ for interior product, so I am wondering can we also define a cohomology for $\iota_X$?
Another question, if $df=0$, then we know locally we can integrate $f$ and get $g=\int^x f$ such that $f=dg$. I am also wondering wether similar formula exists for $\iota_X$, namely: if $\iota_X f=0$, can we locally find $g$ s.t. $f=\iota_X g$?
This is really just a comment, but too long to leave as such. First of all, of course you can define $\ker(\iota_X\colon\Omega^p(M)\rightarrow\Omega^{p-1}(M))/\mathrm{im}(\iota_X\colon\Omega^{p+1}(M)\rightarrow\Omega^p(M))$ for a manifold $M$, because $\iota_X\iota_X=0$. I'd argue this should be called a homology rather than a cohomology (because its differential decreases degree), but that's just a formal issue. Let me denote this temporarily as $H_p^X(M)$.
Although this might seem superficially similar to de Rham cohomology, because you have a differential on the complex of differential forms, I want to stress this is a completely different type of beast. First of all, the de Rham cohomology is natural insofar as it only depends on the smooth structure of the manifold $M$, whereas this "homology" depends on a choice of vector field $X$ and varies drastically depending on that choice (we will see this in a bit). Secondly, the "differential" $\iota_X$ is actually $C^{\infty}(M)$-linear (why? and what does this have to do with the Leibniz rule?), so the "homology" groups $H_p^X(M)$ naturally carry the structure not only of $\mathbb{R}$-vector spaces, but of $C^{\infty}(M)$-modules. This means that they are much richer on one hand, but also that they can be much more complicated on the other hand.
To illustrate my belief that these "homologies" are very complicated, let's consider the arguably easiest case: $M=\mathbb{R}$ and we only look at degree $p=0$. In this case, the "cycles" are $\Omega^0(M)=C^{\infty}(\mathbb{R})$ and the "boundaries" are $\iota_X\Omega^1(\mathbb{R})\subseteq C^{\infty}(\mathbb{R})$, which are all the smooth functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ that can be written in the form $f=\omega(X)$ for some $1$-form $\omega\in\Omega^1(\mathbb{R})$. Let's write $X=a\frac{\partial}{\partial t}$ for a smooth function $a\colon\mathbb{R}\rightarrow\mathbb{R}$ and $\omega=b\,dt$ for a smooth function $b\colon\mathbb{R}\rightarrow\mathbb{R}$, so that $\omega(X)=ab$. Thus, the question of determining $\iota_X(\Omega^1(\mathbb{R}))$ comes down to determining the smooth functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$ for which there exists a smooth function $b\colon\mathbb{R}\rightarrow\mathbb{R}$, such that $f=ab$. This is nothing else than the principal ideal $(a)$ generated by $a$ in $C^{\infty}(\mathbb{R})$ (it is not surprising that the image of $\iota_X$ is an ideal rather than just a subgroup, because $\iota_X$ is $C^{\infty}(\mathbb{R})$-linear). Thus, the "homology" $H_0^X(\mathbb{R})$ is the residue class ring $C^{\infty}(M)/(a)$.
Let's try some examples. The easiest case is when $X$ is the zero vector field, i.e. $a=0$, in which case $H_0^0(\mathbb{R})=C^{\infty}(\mathbb{R})$, which is huge! Perhaps it is reasonable to figure out when $H_0^X(\mathbb{R})=0$? This happens iff $(a)=C^{\infty}(\mathbb{R})$, i.e. iff $a$ is a unit, which happens iff $a$ has no zeroes, i.e. iff $X$ has no zeroes. Interesting.
Now, what happens if we let $X=t^{n+1}\frac{\partial}{\partial t}$? If we can write, $f=t^{n+1}b$, then repeated application of the product rule shows that $f(0)=\dotsc=f^{(n)}(0)=0$. On the other hand, a version of Taylor's theorem implies that we can always write $f=\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}t^k+gt^{n+1}$ with a smooth function $g\colon\mathbb{R}\rightarrow\mathbb{R}$, so if $f(0)=\dotsc=f^{(n)}(0)=0$, then $f=gt^{n+1}$. It follows that $\iota_X(\Omega^1(\mathbb{R}))$ is precisely the ideal of those smooth functions whose derivatives up to order $n$ vanish at $0$. If we consider the surjection $C^{\infty}(M)\rightarrow\mathbb{R}^{n+1},\,f\mapsto(f(0),\dotsc,f^{(n)}(0))$, that's precisely its kernel, so the isomorphism theorem implies $H_0^{t^{n+1}\frac{\partial}{\partial t}}(\mathbb{R})\cong\mathbb{R}^{n+1}$.
What happens if we let $X=(t-p_1)\cdot\dotsc\cdot(t-p_n)\frac{\partial}{\partial t}$, where $p_1,\dotsc,p_n\in\mathbb{R}$ are distinct points? The ideals $(t-p_1),\dotsc,(t-p_n)$ in $C^{\infty}(\mathbb{R})$ are pairwise comaximal, so the CRT implies that $C^{\infty}(\mathbb{R})/\left((t-p_1)\cdot\dotsc\cdot(t-p_n)\right)\cong C^{\infty}(\mathbb{R})/(t-p_1)\times\dotsc\times C^{\infty}(\mathbb{R})/(t-p_n)$. Furthermore, $C^{\infty}(\mathbb{R})/(t-p_1)\cong\mathbb{R}$ via $[f]\mapsto f(p_1)$ (this is similar to the previous paragraph), so in total we get $H_0^{(t-p_1)\cdot\dotsc\cdot(t-p_n)\frac{\partial}{\partial t}}(\mathbb{R})\cong\mathbb{R}^n$.
Now, I've been sweeping something under the rug. Namely, as I said previously the "homologies" $H_p^X(M)$ carry the structure of a $C^{\infty}(M)$-module, but so far I've only been describing $\mathbb{R}$-module structures. Take for example $X=t^2\frac{\partial}{\partial t}$ and $Y=(t^2-1)\frac{\partial}{\partial t}$. We observed that $H_0^X(\mathbb{R})\cong\mathbb{R}^2\cong H_0^Y(\mathbb{R})$, but this is only an isomorphism of $\mathbb{R}$-modules. The $C^{\infty}(M)$-module structure on $H_0^X(\mathbb{R})\cong\mathbb{R}^2$ translates to $h.(x_0,x_1)=(h(0)x_0,h^{\prime}(0)x_0+h(0)x_1)$, whereas the $C^{\infty}(\mathbb{R})$-module structure on $H_0^Y(\mathbb{R})\cong\mathbb{R}^2$ translates to $h.(x_0,x_1)=(h(1)x_0,h(-1)x_1)$. These are not isomorphic as $C^{\infty}(\mathbb{R})$-modules. This is due to the simple fact that the annihilator ideal is invariant under $C^{\infty}(M)$-linear isomorphisms and the annihilator ideal of $C^{\infty}(M)/(a)$ is $(a)$ itself. You can use this and the previous techniques to show that if $a$ is a polynomial, then $H_0^X(\mathbb{R})$ is $\deg(a)$-dimensional as $\mathbb{R}$-vector space, but the $C^{\infty}(\mathbb{R})$-module structure on $H_0^X(\mathbb{R})$ is actually enough to recover $a$ up to multiplication by a non-zero real number.
This should already demonstrate that these "homologies" are both very complicated and rich. Even in this test case, one can ask what happens if $a=e^t-1$ or $a$ is the famous function $a(t)=e^{-1/t}$ for $t>0$ and $a(t)=0$ for $t\le0$. I haven't thought about these questions in detail, but the answers certainly aren't trivial.