I will try to address what I think may be the confusion; if I have misunderstood you, just let me know.
My understanding of your question is that there is a perceived difference in the zero of the intersection and the zero of the sum. This is not the case: there is a single vector space, $$V,$$ and only one zero element of $$V.$$
What we are trying to show is ultimately set-theoretic in nature: we want to show that if an element $$v\in\bigcap_{i=1}^{n}\text{ker}(T_{i}),$$ then $$v\in \text{ker}\left(\sum_{i=1}^{n}T_{i} \right).$$ Let's break down what all this means:
To say that a vector belongs to the kernel of a linear transformation means that the linear transformation sends that vector to zero. So, by assuming $$v\in\bigcap_{i=1}^{n}\text{ker}(T_{i})$$ we are saying that $$v$$ makes all of the linear transformations zero; i.e.
$$T_{1}(v)=0,~ T_{2}(v)=0,\ldots, T_{n}(v)=0$$
In each case above the zero is the zero element of $$V,$$ they are not different zeros. Now, to show that $$v\in \text{ker}\left(\sum_{i=1}^{n}T_{i} \right),$$ we must show that when we plug $$v$$ into the function/linear transformation $$\sum_{i=1}^{n}T_{i}$$ we still get zero. For this we compute:
$$\left( \sum_{i=1}^{n}T_{i}\right)(v)=\sum_{i=1}^{n}T_{i}(v)=T_{1}(v)+T_{2}(v)+\ldots+T_{n}(v)=0+0+\ldots+0=0$$
Hence, $$v\in\text{ker}\left(\sum_{i=1}^{n}T_{i} \right).$$ Since $$v\in\bigcap_{i=1}^{n}\text{ker}(T_{i})$$ was arbitrary, the proof is finished.
Let's take an example to help us along. Let our vector space be
$$V=\mathbb{R}^{3}=\left\{\begin{bmatrix}x\\ y\\ z \end{bmatrix}: x, \, y, \, z\in\mathbb{R} \right\}$$
and our linear transformations $$T_{1}$$ & $$T_{2}$$ be given by
$$T_{1}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}0\\y\\0 \end{bmatrix}$$
and
$$T_{2}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}x\\0\\0 \end{bmatrix}$$
The zero element in this example is $$\begin{bmatrix}0\\0\\0\end{bmatrix}$$
Notice that $$T_{1}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
if and only if $$y=0.$$ Hence, $$\text{ker}(T_{1})=xz$$ plane. Similarly, $$\text{ker}(T_{2})=yz$$ plane. The intersection of these two kernels is the entire $$z$$ axis; i.e.
$$\bigcap_{i=1}^{2}\text{ker}(T_{i})=\left\{\begin{bmatrix}0\\0\\z \end{bmatrix}: z\in\mathbb{R} \right\}$$
(Drawing a picture of the two planes to see that their intersection is the $$z$$ axis may be helpful)
Now, if we wish to demonstrate the general proof you're working on in this example, we take an element $$\begin{bmatrix}0\\0\\z\end{bmatrix}\in \bigcap_{i=1}^{2}\text{ker}(T_{i})$$ and compute using the definitions of $$T_{1}$$ & $$T_{2}$$
$$\left(T_{1}+T_{2}\right)\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)=T_{1}\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)+T_{2}\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)=\begin{bmatrix}0\\0\\0 \end{bmatrix} + \begin{bmatrix}0\\0\\0 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}
$$
Hence, $$\begin{bmatrix}0\\0\\z\end{bmatrix}\in\text{ker}\left(\sum_{i=1}^{2}T_{i} \right)$$ as well. Thus, for this specific example,
$$\bigcap_{i=1}^{2}\text{ker}(T_{i})\subseteq \text{ker}\left(\sum_{i=1}^{2}T_{i} \right),$$
as it should be from your general exercise.
This is a long post, but I hope I have understood and addressed your concern. Let me know if anything is unclear/not quite right.
