my question is the following: given a smooth manifold $M$, take a homotopy of maps $f_{t}:M \rightarrow \mathbb{R}, \quad t \in [0,1]$ such that every $f_{t}$ is a Morse function. Do $f_{0}$ and $f_{1}$ have the same critical points' structure? In that case, is it possible to generalize the result to MorseBott functions?

To see why the answer is yes for the first question use the implicit function theorem. For any $t_0$ you can find $\newcommand{\ve}{\varepsilon}$ $\ve >0$ and smooth maps $$\gamma_1,\dotsc, \gamma_N:(t_0\ve,t_0+\ve)\to M$$ such that for any $t\in (t_0\ve,t_0+\ve)$ the points $\gamma_1(t),\dotsc, \gamma_n(t)\in M$ are pairwise distinct and the critical set of $f_t$ consists precisely of these points. Clearly the index of $\gamma_i(t)$ is independent of $t$, for any $i$. If we denote consider the set $$ C:=\bigl\lbrace (t,x)\in [0,1]\times M;\;\;df_t(x)=0\bigr\rbrace, $$ then the above fact shows that the natural projection $$ C\ni (t,x)\mapsto t\in [0,1] $$ is a covering map. As for the second question, I will give an example. Denote by $\DeclareMathOperator{\Gr}{Gr}$ by ${\Gr}_k(V)$ the Grassmannian of $k$dimensional subspaces of the complex Hermitian space $V$. For any selfadjoint operator $A: V\to V$ we have a smooth function $${\Gr}_k(V)\ni L \mapsto f_A(L)= {\rm Re}\; {\rm tr}(AP_L)\in \mathbb{R}, $$ where $P_L$ denotes the orthogonal projection onto $L$. This function is MorseBott for any $A$, but it is Morse if and only if the eigenvalues of $A$ are simple, i.e., pairwise distinct. You can now easily construct a smooth family of selfadjoit operators $A_t$, $t\in [0,1]$ such that $A_0=\boldsymbol{1}_V$ but $A_t$ has simple eigenvalues for any $t>0$. 

