Let $\Delta$ be the complex with vertices
$$
x, y, a, b_1, b_2, \dots, b_k, c_1, c_2, \dots, c_k
$$
generated by the following $2$-simplices:
$$
(a,b_i,c_i), (a,b_i, y), (b_i, c_i, y), (a, c_i, x)
$$
For any individual $i$, these 2-simplices generate a contractible subcomplex $\Delta_i$ whose realization is $D^2$ because all of the simplices contain $a$. The intersection of the subcomplexes $\Delta_i$ and $\Delta_j$ is the pair of edges $(a,x)$ and $(a,y)$. The geometric realization of $\Delta$ is a union of discs, glued along a common edge. It is contractible and $\Delta$ has Betti numbers $1, 0, 0, \dots$

The quotient complex $\phi \Delta$ has vertices $x, a, b_i, c_i$ and is generated by $2$-simplices
$$
(a,b_i,c_i), (a,b_i, x), (b_i, c_i, x), (a, c_i, x).
$$
Each value of $i$ gives the boundary of a tetrahedron $T_i$. The intersection of $T_i$ with $T_j$ is the edge $(a,x)$. The geometric realization of $\phi \Delta$ is a union of $k$ 2-spheres, glued along a common edge, and its Betti numbers are $1, 0, k, 0, \dots$.

(This is really $k$ copies of the example you described in a now-deleted comment.)

As a result, the Betti numbers of $\phi \Delta$ aren't bounded by those of $\Delta$.

Here is a more general result.

Consider the standard simplex with vertices $(v_0, \dots, v_k)$, and let $K$ and $L$ be subcomplexes of it.

Let $X$ be the subcomplex of $(x, v_0, \dots, v_k)$ generated by the following simplices:

$(x)$

$(v_0, \dots, v_k)$

$(x, c_1, \dots, c_l)$ whenever $(c_1, \dots, c_l)$ is a simplex of $K$

Then $X$ is the mapping cone of the map $K \to \Delta^k$ and
$$
\widetilde H_*(X) \cong \widetilde H_{*-1}(K).
$$

Similarly, let $Y$ be the subcomplex of $(y,v_0,\dots,v_k)$ whose reduced homology groups are a shift of the homology groups of $L$.

Let $\Delta$ be the union of $X$ and $Y$, as a subcomplex of $(x,y,v_0,\dots,v_k)$. The intersection of $X$ and $Y$ is the simplex $(v_0,\dots,v_k)$ which is contractible, and so
$$
\widetilde H_*(\Delta) \cong \widetilde H_{*-1}(K) \oplus \widetilde H_{*-1}(L).
$$
The complex $\phi \Delta$ is the subcomplex of $(x, v_0, \dots, v_k)$ generated by $(x)$, $(v_0,\dots,v_k)$, simplices associated to $K$, and simplices associated to $L$. As a result, it is also a mapping cone: the mapping cone of the map $K \cup L \to \Delta^k$, and so
$$
\widetilde H_*(\phi \Delta) \cong \widetilde H_{*-1}(K \cup L).
$$
Therefore, it suffices to find complexes $K$ and $L$ with not much homology but whose union has large homology.

Now let's do a specific example to show that we can get large growth.

For any $d \geq 0$, let $K$ be the subcomplex generated by the $d$-simplices containing $v_0$, and let $L$ be the subcomplex generated by the $d$-simplices containing $v_1$. Then $K$ linearly contracts onto $v_0$ and $L$ linearly contracts onto $v_1$, so $\widetilde H_*(K) = \widetilde H_*(L) = 0$, and so
$$\widetilde H_*(\Delta) = 0.$$
Also, by the Mayer-Vietoris sequence, we have
$$\widetilde H_*(\phi \Delta) = \widetilde H_{*-1}(K \cup L) = \widetilde H_{*-2}(K \cap L).$$

The complex $K \cap L$ is the union of all of the $(d-1)$-simplices of $(v_2,\dots,v_k)$--the $(d-1)$-skeleton of $\Delta^{k-2}$. Because of this there is a long exact sequence
$$
0 \to \widetilde H_{d-1}(K \cap L) \to C_{d-1}(\Delta^{k-2}) \to \dots \to C_1(\Delta^{k-2}) \to C_0(\Delta^{k-2}) \to \Bbb Z \to 0,
$$
the rank of $H_{d+1}(\phi \Delta)$ is the alternating sum of the ranks, which is the alternating sum of the number of simplices in $\Delta^{k-2}$:

$$
\begin{align*}
b_{d+1} &= \text{rank of }H_{d+1}(\phi \Delta)\\
&= \text{rank of }\widetilde H_{d-1}(K \cap L)\\
&= num\binom{k-1}{d} - \binom{k-1}{d-1} + \binom{k-1}{d-2} - \dots + (-1)^{d-1} \binom{k-1}{1} + (-1)^d \\
&= \binom{k-2}{d}.
\end{align*}
$$
In particular, this is polynomial of degree $d$. If $n = k+3$ is the number of vertices and $i = d+1$, then the growth of the $i$'th Betti number can be $O(n^{i-1})$.