Suppose that $\{v_1,v_2,\dots,v_n\}$ is linearly independent. To show linear independence of $\{v_1,v_1+v_2,v_1+v_2+v_3,\dots,v_1+v_2+\dots+v_n\}$ take a linear combination and set it equal to zero, then prove that the coefficients are necessarily zero. Observe that if,
\begin{align}
a_1(v_1) + a_2(v_1+v_2) + \dots + a_n(v_1+v_2+\dots+v_n) &= 0\\
\Rightarrow \left(\sum_{i=1}^n a_i\right) v_1 + \left(\sum_{i=2}^n a_i\right) v_2 + \dots + a_nv_n &= 0
\end{align}
Which by linear independence of $\{v_1,v_2,\dots,v_n\}$ implies that $\left(\sum_{i=1}^n a_i\right), \left(\sum_{i=2}^n a_i\right),\dots, a_n$ are all equal zero, from which it's easy to see that $a_1,a_2,\dots,a_n$ are all zero as well so we are done.
It is true that two vectors are dependent if they "point in the same (or opposite) direction", i.e. if they are aligned.
But that is not totally true for three vectors in $3$D or more.
In the sense that, when the three vectors are aligned, i.e. parallel, i.e. when they are scalar multiples of each other, they are for sure dependent.
But the definition of linear dependency of three vectors is wider than being parallel: it includes also the case in which they are co-planar, although not parallel.
If you want to see that geometrically, taking the three vectors as position vectors from the origin, if they define a full $3$D parallelepiped then they are independent, if instead the parallelepiped collapses into a flat figure or segment then the vectors are dependent.
Algebraically this translates into the fact whether the matrix formed by the three vectors has full rank ($3$) or less.
Similarly for $n$ vectors of $m$ dimensions.
Then from the theory of linear system you know that, in a homogeneous system, if the matrix has full rank then it has the only solution $(0,0, \cdots, 0)$ which corresponds to the combination coefficients to be all null.
In reply to your comment, in ${\mathbb R}^2$ if you have two non-aligned = independent vectors, then a third one will lie on their same plane (the $x,y$ plane).
In the geometric interpretation, the parallelepiped (the hull) will be flat, i.e. dimension 2, which is less than 3, the number of vectors.
In the algebraic interpretation, a matrix $3 \times 2$ cannot have a rank greater than two: so 3 (or more) 2D vectors are necessarily dependent.
final note (to clarify what might be the source of your confusion)
The (in)dependence of $n$ vectors in ${\mathbb R}^m$ is defined for the whole set of $n$ vectors: they might be dependent, notwithstanding that a few of them ($q<n, \; q\le m$) could be independent. Yet if one is dependent on another (or other two, etc.), then the whole set is dependent.
And in fact it is a common task, given $n$ vectors, to find which among them represent an independent subset: the minor in the matrix with non-null determinant, the larger giving the rank.
Best Answer
$\mathbb R^2$ has dimension $2$, so a set of $3$ vectors from $\mathbb R^2$ can never be linearly independent.
(In your case, the three vectors are even more dependent than they have to be, since they are all parallel).