Your did pick the right vector : $v_pf:=w_{p_i}(f|_{M_i})$. But here you should be more specific about what the function $f|_{M_i}= f \circ \iota_i$ mean. Because there are uncountably many copies of $M_i$ inside $M$, that is we can have many inclusion $\iota_i : M_i \hookrightarrow M$ as $\iota_i(x) = (c_1,\cdots,c_{i-1},x,c_{i+1},\cdots,c_n)$. But we will see that the correct inclusion is $\iota_i(x) = (p_1,\cdots,p_{i-1},x,p_{i+1},\cdots,p_n)$. The other choice of injection is fail because the operator $v_p : C^{\infty}(M) \to \mathbb{R}$ correspond to it, is not a derivation (it fails to satisfy Leibniz Rule). I'll put the proof in the note below.
Now, for any $g \in C^{\infty}(M_i)$ and $w_{p_i}\in T_{p_i}M_i$, the vector $ v_p \in T_pM$ defined as $v_p(f):=w_{p_i}(f \circ \iota_i)$, with $\iota_i(x) = (p_1,\cdots,p_{i-1},x,p_{i+1},\cdots,p_n)$, satisfy
$$
d \pi_i (v_p)g =v_p (g \circ \pi_i) = w_{p_i} (g \circ \pi_i)|_{M_i} = w_{p_i} (g \circ \pi_i \circ \iota_i) = w_{p_i} g.
$$
$\textbf{Note : }$
$\textbf{The vectors } v_{p}f:=w_{p_i}(f \circ \iota_i), \textbf{ where } \iota_i (x):= (c_1,\dots,c_{i-1},x,c_{i+1},\dots,c_n) \textbf{ and } \iota_i(p_i)\neq p = (p_1,\dots,p_n), \textbf{ is not a derivation}$.
Let $f,g \in C^{\infty}(M)$ be arbitrary. By def we have
\begin{align*}
v_p(fg) &= w_{p_i}(fg)|_{M_i} = w_{p_i} ((fg) \circ \iota_i) = w_{p_i} (f|_{M_i}g|_{M_i})\\
&= f(c_1,\dots,c_{i-1},p_i,c_{i+1},\dots,c_n) v_pg + g(c_1,\dots,c_{i-1},p_i,c_{i+1},\dots,c_n) v_pf
\end{align*}
which is not necessarily equal to $f(p)v_pg + g(p) v_pf$.
As @PaulFrost says, we can alternatively prove the surjectivity of $d\pi_{i}$ by find the right-inverse for it. That is a linear map $L : T_{p_i}M_i \to T_pM$ such that $d\pi_i \circ L = \text{Id}_{T_{p_i}M_i}$. The appropriate right-inverse is
$$
d\iota_i : T_{p_i}M_i \to T_pM,
$$
since for any $w_{p_i} \in T_{p_i}M_i, f\in C^{\infty}(M_i)$,
$$d\pi_i \circ d\iota_i (w_{p_i}) f = d(\pi_i \circ \iota_i)(w_{p_i})f = w_{p_i}(f \circ \pi_i \circ \iota_i) = w_{p_i}f.$$
However, the similar construction is used to prove that actually
$T_pM = T_p(M_1 \times \cdots \times M_n) \cong T_{p_1}M_1 \times \cdots \times T_{p_n}M_n$. (See here for example)