It does change - you get an additional term. I worked this out in real time, so there was probably a more elegant way to do this, and there may well be a typo or two. Corrections welcome. I also used the term "Lorentz vector" to refer to what we'd call a 4-vector in (3+1)-dimensions, and the term "spatial vector" to refer to what we'd call a 3-vector.
We'll stick to (2+1)-dimensions, so the general Lorentz boost (acting on a Lorentz vector) can be represented by the following matrix:
$$\Lambda = \pmatrix{
\gamma & -\gamma \beta n_x & -\gamma \beta n_y \\
-\gamma \beta n_x & 1+ (\gamma-1)n_x^2 & (\gamma - 1)n_x n_y \\
-\gamma \beta n_y & (\gamma -1)n_x n_y & 1 + (\gamma-1)n_y^2}$$
where the velocity of the second frame with respect to the first is $\vec \beta = \beta \left(n_x, n_y\right)$.
Without loss of generality, let's align our x-direction with the first boost, so the first Lorentz transformation can be represented by
$$ \Lambda_1 = \pmatrix{
\gamma_1 & -\gamma_1 \beta_1 & 0 \\
-\gamma_1 \beta_1 & \gamma_1 & 0 \\
0 & 0 & 1 }$$
So the composition of boosts would be
$$ \Lambda_{21} = \Lambda_2 \cdot \Lambda_1 = \pmatrix{
\gamma_2 & -\gamma_2 \beta_2 n_x & -\gamma_2 \beta_2 n_y \\
-\gamma_2 \beta_2 n_x & 1+ (\gamma_2-1)n_x^2 & (\gamma_2 - 1)n_x n_y \\
-\gamma_2 \beta_2 n_y & (\gamma_2 -1)n_x n_y & 1 + (\gamma_2-1)n_y^2}\cdot \pmatrix{
\gamma_1 & -\gamma_1 \beta_1 & 0 \\
-\gamma_1 \beta_1 & \gamma_1 & 0 \\
0 & 0 & 1 }$$
$$ = \pmatrix{
\gamma_2 \gamma_1(1+\beta_2 \beta_1 n_x) & -\gamma_2 \gamma_1(\beta_2 n_x + \beta_1) & -\gamma_2 \beta_2 n_y \\
-\gamma_2 \gamma_1 n_x(\beta_2 + n_x \beta_1) + \gamma_1\beta_1(n_x^2-1) & \gamma_2 \gamma_1 n_x(\beta_2 \beta_1 + n_x) + \gamma_1(1-n_x^2) & (\gamma_2-1)n_x n_y \\
-\gamma_2\gamma_1 n_y(\beta_2 + n_x \beta_1) + \gamma_1 \beta_1 n_x n_y & \gamma_2 \gamma_1 n_y(\beta_2 \beta_1 + n_x) - \gamma_1 n_x n_y & 1 + (\gamma_2-1)n_y^2 }$$
If $n_x=1$ and $n_y=0$, this reduces to
$$\Lambda_{21} = \pmatrix{
\gamma_2 \gamma_1(1+\beta_2 \beta_1) & -\gamma_2 \gamma_1 (\beta_2 + \beta_1) & 0 \\ -\gamma_2 \gamma_1 (\beta_2 + \beta_1)& \gamma_2 \gamma_1(1+\beta_2 \beta_1) & 0 \\ 0 & 0 & 1}$$
You can see that if we define $\gamma_{21} = \gamma_2 \gamma_1 (1+\beta_2 \beta_1)$ and $\beta_{21} = \frac{\beta_2 + \beta_1}{1+\beta_2 \beta_1}$ then we just have another boost:
$$ \Lambda_{21} = \pmatrix{
\gamma_{21} & -\gamma_{21} \beta_{21} & 0 \\
-\gamma_{21} \beta_{21} & \gamma_{21} & 0 \\
0 & 0 & 1 }$$
So an object at rest in the original frame would be seen with Lorentz velocity
$$ u = \pmatrix{
\gamma_{21} & -\gamma_{21} \beta_{21} & 0 \\
-\gamma_{21} \beta_{21} & \gamma_{21} & 0 \\
0 & 0 & 1 } \cdot \pmatrix{ 1 \\ 0 \\ 0} = \pmatrix{ \gamma_{21} \\ -\gamma_{21} \beta_{21} \\ 0} $$
and spatial velocity $-\vec \beta_{21}$ after the two frame changes.
In general, sadly, things are more complicated. We would have that
$$ u = \pmatrix{\gamma_2 \gamma_1(1+\beta_2 \beta_1 n_x) \\ -\gamma_2 \gamma_1 n_x(\beta_2 + n_x \beta_1) - \gamma_2\beta_1n_y^2 \\ -\gamma_2\gamma_1 n_y(\beta_2 + n_x \beta_1) + \gamma_1 \beta_1 n_x n_y}$$
where we've used that $n_x^2-1=-n_y^2$.
We can do a bit more rearranging to get
$$ u = \pmatrix{\gamma_2 \gamma_1(1+\beta_2 \beta_1 n_x) \\ -\gamma_2 \gamma_1 (n_x\beta_2 + \beta_1) + n_y^2 \gamma_1 \beta_1(\gamma_2 - 1) \\ -\gamma_2\gamma_1 n_y \beta_2 - n_x n_y \gamma_1 \beta_1(\gamma_2 - 1)}$$
Getting closer. Recall that $\gamma^2 - 1 = \beta^2\gamma^2$, so $\gamma-1 = \frac{\beta^2\gamma^2}{\gamma+1}$, and we can say
$$ u = \pmatrix{\gamma_2 \gamma_1(1+\beta_2 \beta_1 n_x) \\ -\gamma_2 \gamma_1 (n_x\beta_2 + \beta_1) + \frac{n_y^2 \gamma_2^2\gamma_1 \beta_1\beta_2^2}{\gamma_2+1} \\ -\gamma_2\gamma_1 n_y \beta_2 - \frac{n_x n_y \gamma_2^2\gamma_1 \beta_2^2\beta_1}{\gamma_2+1}}$$
Letting $\gamma_{21} = \gamma_2 \gamma_1 (1+\beta_2 \beta_1 n_x) = \gamma_2 \gamma_1 (1 + \vec \beta_2 \cdot \vec \beta_1)$ and factoring it out,
$$ u = \gamma_{21} \pmatrix{1 \\ -\frac{n_x \beta_2 + \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1} + \frac{\gamma_2}{\gamma_2 + 1} \frac{n_y^2 \beta_2^2 \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1} \\ -\frac{n_y \beta_2}{1 + \vec \beta_2 \cdot \vec \beta_1} - \frac{\gamma_2}{\gamma_2+1} \frac{n_x n_y \beta_2^2 \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1}}$$
Let's define $\vec \beta_A = \frac{\vec \beta_2 + \vec \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1}$ and $\vec \beta_B = \frac{\gamma_2}{\gamma_2+1}\frac{\beta_2 \times (\beta_2 \times \beta_1)}{1 + \vec \beta_2 \cdot \vec \beta_1} = \frac{\gamma_2}{\gamma_2+1}\frac{\vec \beta_2 (n_x \beta_2 \beta_1) - \vec \beta_1 (\beta_2^2)}{1 + \vec \beta_2 \cdot \vec \beta_1}$. Note that
$$\hat x \cdot \vec \beta_A = \frac{n_x \beta_2 + \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1}$$
$$ \hat y \cdot \vec \beta_A = \frac{n_y \beta_2}{1 + \vec \beta_2 \cdot \vec \beta_1}$$
$$ \hat x \cdot \vec \beta_B = \frac{\gamma_2}{\gamma_2+1}\frac{n_x^2 \beta_2^2\beta_1 - \beta_2^2\beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1} = \frac{\gamma_2}{\gamma_2+1}\frac{-n_y^2 \beta_2^2 \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1} $$
and
$$ \hat y \cdot \vec \beta_B = \frac{\gamma_2}{\gamma_2+1}\frac{n_x n_y \beta_2^2 \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1}$$
So at last, we can define our velocity addition formula. An object at rest in the original frame will be observed to have Lorentz velocity
$$ u = \pmatrix{\gamma_{21} \\ -\gamma_{21} \vec \beta_{21} }$$
and spatial velocity
$$ -\vec \beta_{21} = -\left(\vec \beta_A + \vec \beta_B\right) = -\left(\frac{\vec \beta_2 + \vec \beta_1}{1 + \vec \beta_2 \cdot \vec \beta_1} + \frac{\gamma_2}{\gamma_2 + 1} \left[\frac{\vec \beta_2 \times (\vec \beta_2 \times \vec \beta_1)}{1 + \vec \beta_2 \cdot \vec \beta_1}\right] \right)$$
where $\gamma_{21} = \gamma_2 \gamma_1 (1 + \vec \beta_2 \cdot \vec \beta_1)$.
Ugh. Gross. But there you have it. Obviously if the velocities are colinear then the second term vanishes, but in general it alters both the magnitude and the direction of the resulting velocity. Also note that the order matters - flipping the order of the boosts changes the second term in a non-trivial way (not just a sign flip).
