Does thermal expansion and expending a solid object depends on processes? As we know that for a rod of length $l_o$ if we increase the temprature by $2T$, the final length of the rod will be,
$$l_{f_1}=l_o(1+2\alpha T)$$
Where as if we first increase to T and then stop,
$$l=l_o(1+\alpha T)$$
And then again increase the temprature by T the initial length will now be $l$,
$$l_{f_2}=l(1+\alpha T)=l_o(1+\alpha T)^2$$
We can clearly see $l_{f_1}≠l_{f_2}$.
We can say it to be equal approximately for small change in temprature, but for a significant change what would you say? As $\alpha$ will have a small change due to increase in temprature, it can't counterbalance both lengths.
Same in the case of manually stretching a rod by $2x$. Workdone will not be same in expanding from $0$ to $2x$ and $0$ to $x$ then again expanding $x$. As,
$$Y=\frac{Fl_o}{Ax}$$ Workdone will be $$W=\frac{kx^2}{2}$$
Where $k=\frac{YA}{l_o}$, and then same process.
What would you explain?

$Y$ = Young's modulus
$\alpha$ = coefficient of linear thermal expansion
$F$ = Applied force to expand rod

 A: The equation for the expansion of a rod:
$$ \ell(T) = \ell_0(1 + \alpha \Delta T) \tag{1}$$
is only an approximation. The actual equation will be complicated and will depend on the material, however any equation can be approximated by a Taylor series. In this case we have some function for the length of the rod $\ell(T)$ and we use a Taylor expansion about $T_0$:
$$ \ell(T) = \ell_0 + \frac{d\ell}{dT}(T_0) (T - T_0) + O(T-T_0)^2 $$
where we are expanding about the temperature $T_0$ and the rod length at this temperature is $\ell_0 = \ell(T_0)$. If we define:
$$\begin{align}
\Delta T &= T - T_0 \\
\alpha(T_0) &= \frac{1}{\ell_0} \frac{d\ell}{dT}(T_0)
\end{align}$$
and neglect the quadratic and higher terms then we get equation (1):
$$ \ell(T) = \ell_0(1 + \alpha(T_0) \Delta T)$$
except that I have explicitly shown that the parameter $\alpha(T_0)$ is a function of our reference temperature $T_0$ and not a constant.
And this provides the answer to your question. If you consider the expansion in two stages i.e. from $T_0$ to $T_1$, and then from $T_1$ to $T_2$, the value of $\alpha$ would have to be different for the second expansion because it would be $\alpha(T_1)$ instead of $\alpha(T_0)$.
In practice the dependence of $\alpha$ on temperature is small so we treat it as constant as long as the temperature range isn't too large. In that case we get for two successive changes $\Delta T$:
$$ \ell = \ell_0(1 + \alpha\Delta T)^2 = \ell_0(1 + 2\alpha\Delta T + \alpha^2\Delta T^2) \approx \ell_0(1 + 2\alpha\Delta T) $$
where we assume that since $\alpha$ is small the quadratic term is so small it can be ignored.
A: Your starting equation is only an approximation for small temperature changes.  The exact relationship is $$\frac{dl}{dT}=\alpha l$$
For cases in which both force and temperature change, both effects need to be taken into account as follows:  $$dF=YA\left(\frac{dl}{l}-\alpha dT\right)$$
