Sándor is exactly right, this generally isn't true. However, there is a related statement that is always true, instead of blowing up strict transforms of $N_1$ and $N_2$, you should blow up total transforms.

## Blowing up total transforms

Suppose that $I_1$ and $I_2$ are the ideals defining $N_1$ and $N_2$ in $M$. Let $$Y_1 = Bl_{I_1} M = Bl_{N_1} M.$$
Now define the *total transform of* $I_2$, denoted $\overline{I_{2}}$, to be the ideal formed by extending $I_2$ to $Y_1$, in other words $\overline{I_2} = I_2 \cdot O_{Y_1}$ (note that by the universal property of blowing up $I_1 \cdot O_{Y_1}$ is an invertible sheaf).

$\overline{I_2}$ need not define a manifold, or even a reduced scheme, but we can still blow it up. Set $$Y_{1,2} = Bl_{\overline{I_2}} Y_1.$$
I claim:

**Theorem:** *We have* $Y_{1,2} = Y_{2,1}$ *where the second object is obtained in the same way but blowing up* $I_2$ *first.*

There are at least two ways to see this.

- You can do this from the universal property of blowing up.
- Since you already assumed that $M$ is a manifold, it is integral. Thus the charts of a blowup can be computed as follows. Suppose $M = \text{Spec } A$ is affine for simplicity. If $\langle x_1, \dots, x_n\rangle$ generate an ideal $I$, then as $i = 1,...,n$ varies, $Y_{I,i} = \text{Spec } A[x_1/x_i, \dots, x_i/x_i, \dots, x_n/x_i]$ are affine charts covering the blowup. $Y_I = Bl_I Y$.

**EDIT:** As Dustin Cartwright points out in a comment below, blowing up $I_1$ followed by blowing up $\overline{I_2}$ is the same as blowing up $I_1 \cdot I_2$.

## A normal crossings example

I'd like to give you an example in $\mathbb{A}^4 = \text{Spec } k[x,y,u,v]$ showing the difference between blowing up strict transforms and total transforms, even when the $N_i$ intersect transversally (are in normal crossings). Let $N_1 = V(x,y)$ be a plane and let $N_2 = V(y,u,v)$ be a line.

Suppose we blowup the line $N_2$ first, then the strict transform and the total transform of $N_1$ coincide. In particular, $Y_{2,1}$ is the object you were considering.

On the other hand, let's blow up $N_1$ first, in this case the strict transform $\widetilde{N_2}$ of $N_2$ is a line, but the total transform $\overline{N_2}$ of $N_2$ is two lines. $\overline{N_2}$ contains the strict transform but also one of the new $\mathbb{P}^1$'s lying over the origin of $N_1 \subseteq \mathbb{A}^4$. The blowups of the strict and total transform of $N_2$ on $Y_1$ are thus manifestly different.

One can check that $Y_{1,2} = Y_{2,1}$ if you did the total transform blowup, and so blowing up the strict transforms does not commute.

**Note:** If I recall correctly, you can also do the example when two planes are *kissing.* $N_1 = V(x,y)$, $N_2 = V(u,v)$. The same sort of thing happens, but it's even messier. I've seen this example in various papers on resolution of singularities.

## In the literature

The difference between strict and total transforms plays a key role in modern resolution of singularities algorithms. Even though conceptually strict transforms are much easier to think about, they are much *harder* to compute or control. Thus modern algorithms tend to use total transforms instead in *SOME* places, and then peel off any copies of the exceptional you can, while possibly leaving embedded components.

See for example:

- Section 7 of This paper by Bravo-Encinas-Villamayor
- Example 2.3 of This paper by Howard Thompson