What I say below in my answer is what I can say without having read Acemoglu and Autor's article (unfortunately I haven’t it), an answer based on what you report in your question, of course.
I think your graph is correct.
$0 < I_L < I_H < 1$; this condition always holds.
This condition implies that you have to consider only the part of the graph where $ 0 < I_L < 1$ and $0 < I_H < 1$, that is the part in the blue dotted square in the graph below, the $(0,1)\times (0,1)$ square:
Moreover, I have added the 45° green line, as the condition quoted above says that it is always $I_H>I_L$, so that the graph should be comprised in the part of the plane below the 45° green line.
This condition is respected in your graph, in the $(0,1)\times (0,1)$ square.
Instead, this condition is not respected in the graph of Acemoglu and Autor's paper you posted.
As I said, I didn't read the paper of Acemoglu and Autor but, on the basis of what you refer, I suppose there is an inaccuracy in the graph.
According to the condition $I_L < I_H $, the values of the functions for $I_H=0$ can't be strictly positive.
Indeed, in your graph you have a $0$ value for the blue function and a negative value for the red function.
As I show also below, it seems to me that using functions like those in your graph, everything fits.
$$***$$
How can Acemoglu and Autor's graph be qualitatively obtained?
We have, in addition, the assumption:
The paper does not exactly define $\alpha$. However, what it says is
that $\alpha_L (i) / \alpha_M (i)$ and $\alpha_M (i) / \alpha_H (i)$
are continuously differentiable and strictly decreasing, with $i \in
[0,1]$.
How can we obtain a qualitative graph of the two functions, without assuming a particular function for the $\alpha_j(i)$?
The graph of the two functions can be qualitatively depicted evaluating $I_L$ when $L_H=0$, and calculating the derivatives and evaluating their signs.
I re-write below the two equations of Acemoglu and Autor in implicit form, to use them in the following:
\begin{equation}
f_1(I_L, I_H)=\frac{A_M \alpha_M (I_H) M}{I_H - I_L} - \frac{A_H \alpha_H (I_H) H}{1- I_H}=0
\tag{1}\end{equation}
\begin{equation}
f_2 (I_L, I_H)=\frac{A_L \alpha_L (I_L) L}{I_L} - \frac{A_M \alpha_M (I_L) M}{I_H - I_L}=0
\tag{2}\end{equation}
Moreover, it could be useful to write the equations in terms of $I_L$ and $I_H$ as you did in your question:
\begin{equation}
I_L = I_H - \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H)
\tag {1'}\end{equation}
\begin{equation}
I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_L (I_L)}{\alpha_M (I_L)} \frac{M}{L} I_L
\tag {2'}\end{equation}
I suppose, from your question, that the variables are all positive, so we can see from $(1')$ that for $I_H=0$ $I_L$ assumes a negative value, as in your graph (the red fuction).
From $(2')$ it can be easily seen that the function starts from the origin (the blue function) (of course, then, we must exclude the origin as by assumption $I_L\neq0$ and $I_H\neq 0$ and of course equation $(2')$ is equivalent to $(2)$ for $I_L\neq 0$ only, and for $I_L=0$ function $(2)$ is not defined).
Now, let's calculate the derivatives, and evaluate their signs.
For equation $(2)$ I use the implicit function theorem. I re-write it as:
\begin{equation}
f_2 (I_L, I_H)= A_L \frac {\alpha_L(I_L)}{a_M(I_L)} L(I_H-I_L) - A_M M I_L=0
\tag{2''}\end{equation}
From $(2'')$ we can calculate the derivative ($I_{L2}$ is the function implicitly defined by $(2)$ or $(2'')$)
$$\frac{dI_{L2}}{dI_H}=- \frac { \frac {\partial f_2}{\partial I_H}} {\frac{\partial f_2}{\partial I_L}}= \frac {- A_L \frac {\alpha_L(I_L)}{a_M(I_L)}L}{-A_L \frac {\alpha_L(I_L)}{_M(I_L)}L +(I_H-I_L)A_L L\left (\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'-A_M M}>0 $$
where $\left(\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'$ is the derivative of the ratio with respect to $I_L$, which is negative by assumption.
Analogously, we can calculate the derivative for equation $(1)$ using the implicit function theorem, or simply differentiating $(1')$, obtaining that its sign is positive.
In conclusion, we can draw a graph similar to your graph.
$$***$$
The point left to understand is whether the two functions cross in the $(0,1)\times (0,1)$ square, that is an equilibrium exists (in your particular example, yes).
We can resort to the intermediate value theorem, applied to the difference function $g(I_H)$ between the two functions defined by $(1)$ and $(2)$ above, on the interval $[0,1]$. That is, define the function:
$$g(I_H)= I_{L1}(I_H)- I_{L2}(I_H)$$
where $I_{L1}$ and $I_{L2}$ are the functions implicitly defined by $(1)$ and $(2)$, (and explicitated in $(2')$ and $(1')$), that is respectively the red and blue functions in the graph.
For $I_H=0$ we have already seen that $g(I_H) <0$.
For $I_H=1$ we can show that $g(I_H)>0$.
Indeed, from $(1')$ it is easily seen that for $I_H=1$ the function $I_{L1}=1$ (the red function), as also in your graph.
From equation $(2')$, we can see that, for $I_H=1$, $I_{L2}<1$ (the blue function) (I omit the details otherwise this post becomes cumbersome).
Therefore, $g(I_H)$ is positive for $I_H=1$ .
As $g(I_H)$ is a continuous function, as difference of two continuous functions, we can conclude, from the intermediate value theorem, that there must be a point in $(0,1)$ where $g(I_H)=0$, that is where the two red and blue functions in the graph cross.
