Example problem. Average number of actions for time period $x = 248$, average number of actions for time period $y = 310$. Percentage change is $(310-248)/248 = 25\%$ increase.
The actions can be broken into two subgroups; $A$ and $B$.
Average number of $A$-actions for time period $X$ is $114$, average number of $A$-actions for time period $Y$ is $122$. Percentage change is $(122-114)/114 = 7\%$ increase.
Average number of $B$-actions for time period $X$ is $134$, average number of $B$-actions for time period $y$ is $188$. Percentage change is $(188-134)/134 = 42\%$ increase.
action $A$, time period $X$ plus action $B$, time period $X$ is $114 + 134 = 248$
action $B$, time period $Y$ plus action $B$, time period $Y$ is $122 + 188 = 310$
But
percentage change of action $A$ plus percentage change of action $B$ is $7\%+42\%=49\%$ and not the Percentage change of actions $A$ and $B$ together. Why is this?
First, you have to remember that the overall increment is some kind of mean of the individual increment. The overall increment is not the sum of the individual increments.
Let's put some algebra on your example:
$a,c$ are number of $A$-actions in time periods $X$ and $Y$ respectively.
$b,d$ are number of $B$-actions in time periods $X$ and $Y$ respectively.
So, the increment in $A$, $r_A=\dfrac{c-a}a$. Respectively the increment in $B$, $r_B=\dfrac{d-b}b$.
The sum of these increments (which is not the same as the overall increment) is: $r_A+r_B=\dfrac{c-a}a+\dfrac{d-b}b$.
On the other hand, lets $p_A = \dfrac a{a+b}$ and $p_B = \dfrac b{a+b}$ where $p_A+p_B=1$, and let's calculate the weighted mean of $r_A$ and $r_B$: \begin{align} p_Ar_a+p_Br_B&=\frac a{a+b}\cdot\frac{c-a}a+\frac b{a+b}\cdot\frac{d-b}b \\&=\frac{c-a}{a+b}+\frac{d-b}{a+b} \\&=\frac{(c-a)+(d-b)}{a+b} \\&=\frac{(c+d)-(a-b)}{a+b} \end{align}
The overall increment of both $A$-actions and $B$-actions is $\dfrac{(c+d)-(a+b)}{a+b}$.
Let's see when both increments could be equal: \begin{align} r_A+r_B &= p_Ar_A+p_Br_B \\ r_A+(1-p_B)r_B &= p_Ar_A \\ r_A+p_Ar_B &= p_Ar_A \\ r_A &= p_Ar_A-p_Ar_B \end{align} Note that $0\le p_A\le1$ (equality only if $a=0$ or $b=0$ respectively), so in any non-trivial case $a\ne0$ and $b\ne0$, then $0<p_A<1$.
For a positive increment $r_A$, then $r_A>p_Ar_A$. If $r_B$ is also positive, then $p_Ar_A>p_Ar_A+p_Ar_B$ therefor $r_A>p_Ar_A-p_Ar_B$ and $r_A\ne p_Ar_A-p_Ar_B$. A similar analysis will hold if both increments are negative. So the only chances that the sum of increments are equal to the overall increment are certain special cases in which increments cancel eachother.
In the general case overall increment is not the sum of increments.