On a given knot sequence, the b-spline basis functions of a given degree do form a partition of unity, roughly speaking.
Here are the details ...
To construct b-spline basis functions of degree $m$ (or order $k=m+1$)
we start with a sequence of knots $t_0, \ldots, t_{n+k}$.
The b-spline basis functions of degree $m$ on this given
knot sequence are defined recursively.
We start with
$$
\phi^{0}_i(t) =
\begin{cases}
1 & \text{if $t_i \le t < t_{i+1}$} \\
0 & \text{elsewhere}
\end{cases}
$$
Then, for $m \ge 1$ and $i=0,1,\ldots,n$
$$
\phi^{m}_i(t) =
\frac{t - t_i} {t_{i+m} - t_i}\phi^{m-1}_i(t) +
\frac{t_{i+m+1} - t} {t_{i+m+1} - t_{i+1}}\phi^{m-1}_{i+1}(t)
$$
Then we have
\begin{align*}
\phi^{m}_i(t) &= 0 \quad \text{for} \;\; t_0 \le t \le t_{i} \\
\phi^{m}_i(t) &> 0 \quad \text{for} \;\; t_i < t < t_{i+k} \\
\phi^{m}_i(t) &= 0 \quad \text{for} \;\; t_{i+k} \le t \le t_{n+k}
\end{align*}
So $\phi^{m}_i$ is a "hat" shaped function that is non-zero
over the $k$ spans where $t_i < t < t_{i+k}$ and zero elsewhere.
The $\phi^{m}_i$ form a partition of unity on the interval $[t_m,t_{n+1}]$:
$$
\sum_{i=0}^n \phi^{m}_i(t) = 1 \qquad \text{for} \;\; t_{m} \le t \le t_{n+1}
$$
Note that this holds only on the interval $[t_m,t_{n+1}]$.
So, what's wrong with your reasoning?
In your case, you are using knots $\{0,1,2\}$, and you're constructing basis functions with degree $m=1$ (order $k=2$). So, we have $n+k = 2$, and so $n=0$. So, we can construct one basis function $\phi^1_0$, and the theory tells us that it forms a partition of unity on the interval $[t_m,t_{n+1}] = [t_1,t_1] = [1,1]$. This is true, in some vacuous sense, but it's not very interesting.
If you want to define more interesting basis functions of degree 1 on the interval $[0,2]$, then you'll need more knots -- at least one more to the left of $0$, and one more to the right of $2$. You might use the knot sequence $\{0,0,1,2,2\}$, for example. Then you'll get three basis functions of degree $1$ defined on $[0,2]$ and they will again form a partition of unity.