Explaining Ito formula to an analyst From the point of view of analysis, what is Ito formula?
Is it an integral by substitution, or, a radon-nikodym derivative?
Define the probability space 
$$
\left(C\left(\Bbb R_+\right),\sigma\left(C\left(\Bbb R_+\right)\right),P\right),
$$
where $P$ is the standard Brownian motion measure.
Let $f(x)=x^2$, with Ito formula, I write
$$
\int_{C\left(\Bbb R_+\right)} f(X_t)dP(X)=\int_{C\left(\Bbb R_+\right)} \left\{\int_0^t 2X_sdX_s+t \right\} dP\left(X\right).
$$
The previous equation is my heuristic to explain Ito formula to an analyst.
This heuristic is itself inspired by the following heuristic which I once heard

Ito formula is a way of expressing how the measure $P$ changes from $X_t$ to $f(X_t)$

 A: Itō formula has nothing to do with a probability measure. Let me qualify this statement. While stochastic integration ostensibly needs to be performed with the help of a probability measure, the result of stochastic integration is the same for all equivalent measures. In this sense stochastic integration, and therefore also the Itō formula, is a pathwise concept. It cannot be defined pathwise (in general) but the result is pathwise unique (up to indistinguishability). I hōpe this helps :-).
Edit for Bananach: The result of stochastic integration is the same under all equivalent measures: see Proposition 3.6.20 in
Bichteler, Klaus, Stochastic integration with jumps, Encyclopedia of Mathematics and Its Applications. 89. Cambridge: Cambridge University Press. xiii, 501 p. (2002). ZBL1002.60001.
A: It was introduced to me as "the chain rule of Itō calculus" which I think is a quite elegant description, because it explains it's significance and basic applications. If you know how to calculate the derivatives of your functions $f, g$ then you can derive $f \circ g$.
If you know how to derive a stochastic process $W_t$ and a function $h$, then you can use Itō's formula to derive $h(W_t)$.
A: As already pointed out, Itō's formula is a type of chain rule. 
Let $W_t$ be a Brownian motion (with no drift, starting in 0) Naively, you would guess that the following simple chain rule from analysis holds,
$$
df(W_t)=f'(W_t)dW_t.
$$
However, integrating and taking expectations on both sides would then yield
$$
E[f(W_t)]-f(0)=0\quad\forall t\geq 0
$$
(I used that $E[\int_{0}^{t}g(s)dW_s]=0$ for any adapted process $g(s)$. Intuitively, this holds since in the left-endpoint-rule approximations of the integral, each term has zero expectation because $E[W(s)-W(r)]=0$)
This cannot be true in general. For example, if $f(x)=x^2$ as in your question, this would imply that for any $t\geq 0$ you have $W_t=0$ almost surely, which clearly does not hold.
The above motivates an extra term in the chain rule. The correct form, $\frac{1}{2}f''(W_t)dt$, of this extra term at least makes sense in the above example: Since the Wiener process spreads out, the expected value $E[f(W_t)]$ should be increasing for small $t$ when the second derivative $f''(W_0)$ is positive, and decreasing when it is negative.
To arrive at the exact form of Itō's formula (still not rigorously though) one can use a formal Taylor expansion,
$$
f(W_t)-f(W_0)=f'(W_0)dW_t+\frac{1}{2}f''(W_0)(dW_t)^2+\mathcal{O}(dW_t^3),
$$
together with the intuitive insight that "$(dW_t)^2=dt$" (the variance of $W_t-W_0$ is $t$ by definition of the Wiener process) and "$|dW_t|^r=dt^{r/2}$" for $r>2$, i.e. higher moments are negligible as $dt\to 0$ (again, use that $dW_t$ has normal distribution).
Maybe one can paraphrase the last paragraph as follows:
Itō's formula is the chain rule that you get when higher order terms in the Taylor expansion of $f(X_t)$ become important due to excessive variations of $X_t$. 
