Finding a basis for symmetric $k$-tensors on $V$
I assume that $e_1,\dots,e_n$ is a basis of $V$ and that $e^1,\dots,e^n$ the associated dual basis of $V^*$.
First, let's consider the case of arbitrary (not necessarily symmetric) tensors. We note that, by linearity, $$ T(v^{(1)}, \dots, v^{(k)}) = T\left( \sum_{i=1}^n v^{(1)}_i e_i, \dots, \sum_{i=1}^n v^{(k)}_i e_i \right) = T\left( \sum_{i_1=1}^n v^{(1)}_{i_1} e_i, \dots, \sum_{i_k=1}^n v^{(k)}_{i_k} e_{i_k} \right) = \\ \sum_{i_1=1}^n \cdots \sum_{i_k=1}^n v^{(1)}_{i_1} \cdots v^{(k)}_{i_k} T\left(e_{i_1}, \dots, e_{i_k} \right) $$ Now, define the tensor $\tilde T$ by $$ \tilde T = \sum_{i_1=1}^n \cdots \sum_{i_k=1}^n T\left(e_{i_1}, \dots, e_{i_k} \right) e^{i_1} \otimes \cdots \otimes e^{i_k} $$ Prove that $\tilde T(v^{(1)},\dots,v^{(k)}) = T(v^{(1)},\dots,v^{(k)})$ for any $v^{(1)},\dots,v^{(k)}$. That is, $\tilde T = T$. We've thus shown that any (not necessarily symmetric) $k$-tensor can be written as a linear combination of $e^{i_1} \otimes \cdots \otimes e^{i_k}$.
The same applies for symmetric tensors. However, if $T$ is symmetric, then $$ T\left(e_{i_1}, \dots, e_{i_k} \right) = T\left(e_{\sigma(i_1)}, \dots, e_{\sigma(i_k)} \right) $$ for any permutation $\sigma$. Thus, we may regroup the above sum as $$ T = \tilde T = \sum_{i_1=1}^n \cdots \sum_{i_k=1}^n T\left(e_{i_1}, \dots, e_{i_k} \right) e^{i_1} \otimes \cdots \otimes e^{i_k} = \\ \sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \; \frac 1{\alpha(i_1,\dots,i_k)}\sum_{\sigma \in S_k} T\left(e_{\sigma(i_1)}, \dots, e_{\sigma(i_k)} \right) e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)} = \\ \sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \; \frac 1{\alpha(i_1,\dots,i_k)}\sum_{\sigma \in S_k} T\left(e_{i_1}, \dots, e_{i_k} \right) e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)} = \\ \sum_{1 \leq i_1 \leq \cdots \leq i_k \leq n} \frac 1{\alpha(i_1,\dots,i_k)} T\left(e_{i_1}, \dots, e_{i_k} \right) \underbrace{\sum_{\sigma \in S_k} e^{\sigma(i_1)} \otimes \cdots \otimes e^{\sigma(i_k)}}_{\text{basis element for } Sym^k(V)} $$ Thus, we have expressed $T$ as a linear combination of the desired basis elements.
${\alpha(i_1,\dots,i_k)}$ counts the number of times any element $(\sigma(i_1),\dots,\sigma(i_n))$ appears in the summation over $\sigma \in S_n$. As the comment below points out, we have $$ \alpha(i_1,\dots,i_k) = m_1! \cdots m_n! $$ where $m_j$ is the multiplicity of $j \in \{1,\dots,n\}$ in the tuple $(i_1,\dots,i_k)$.
Let's consider $\mathbb{R}^2$ with standard basis $e_1,e_2$.
If $T$ is a symmetric tensor $T:\mathbb{R}^2\times\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$, then we can group basis vectors $e^{i_1}\otimes e^{i_2}\otimes e^{i_3}$ of the tensor power $(\mathbb{R}^2)^{\otimes 3}$ according to whether or not $T$ must send them to the same value:
$$ \begin{array}{rrrr} e_1\otimes e_1\otimes e_1 \\ \hline e_1\otimes e_1\otimes e_2 & e_1\otimes e_2\otimes e_1 & e_2\otimes e_1\otimes e_1 \\ \hline e_1\otimes e_2\otimes e_2 & e_2\otimes e_1\otimes e_2 & e_2\otimes e_2\otimes e_1 \\ \hline e_2\otimes e_2\otimes e_2 \end{array} $$
In other words, the values $T$ takes on any tensor can be determined as long as we know what values $T$ takes on
- $e_1\otimes e_1\otimes e_1$,
- $e_1\otimes e_1\otimes e_2$,
- $e_1\otimes e_2\otimes e_2$,
- $e_2\otimes e_2\otimes e_2$.
These are precisely the basis elements $e_{i_1}\otimes e_{i_2}\otimes e_{i_3}$ with $i_1\le i_2\le i_3$.
Conversely, given any four values $a,b,c,d$ we can arrange for $T$ to take these values on the above basis vectors by writing out
$$ \begin{array}{lll} T & = & a(e^1\otimes e^1\otimes e^1) \\ &+ & b(e^1\otimes e^1\otimes e^2+e^1\otimes e^2\otimes e^1+e^2\otimes e^1\otimes e^1) \\ & + & c(e^1\otimes e^2\otimes e^2+e^2\otimes e^1\otimes e^2+e^2\otimes e^2\otimes e^1) \\ & + & d(e^2\otimes e^2\otimes e^2), \end{array} $$
Generalize.