The short answer is that your presentation is isomorphic to the (slightly more) "standard" presentation - so you computed a correct answer.
More detail -
According to Wolfram and using $K$ to denote the Klein bottle, we have $\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$ whereas you have (according to Aaron Mazel-Gee's comment) $\pi_1(K)\cong \langle a,b\rangle /abab^{-1}$.
The question is, then, are these isomorphic?
Well, we have $cdc^{-1}d = e$ so, taking the inverse of both sides gives $d^{-1}cd^{-1}c^{-1} = e$. But this has the same form as the relation between $a$ and $b$, so now the isomorphism is clear: we map $a$ to $d^{-1}$ and $b$ to $c$.
That is, define $f:\langle a,b \rangle /abab^{-1}\rightarrow \langle c,d\rangle/cdc^{-1}d$ by $f(a) = d^{-1}$ and $f(b) = c$.
I claim this is well defined, for \begin{align*} f(abab^{-1}) &= f(a)f(b)f(a)f(b)^{-1} \\\ &= d^{-1}cd^{-1}c^{-1} \\\ &= (cdc^{-1}d)^{-1} \end{align*} as it should. (Technically, I'm defining $f$ on $\langle a,b\rangle$ and proving it descends to the quotient.)
Finally, rather than show this is 1-1 and onto, instead, show that $g:\langle c,d\rangle/ cdc^{-1}d\rightarrow \langle a,b\rangle/ abab^{-1}$ defined by $g(c) = b$ and $g(d) = a^{-1}$ is well defined, and the inverse to $f$, so $f$ is the desired isomorphism.
The two spaces $K$ and $T$ are both path connected and so does $K\vee T$, since any two point $a\in K, b\in T$ and be connected by a composite path $f*g$ from $a$ to $b$ where $f(0)=a,f(1)=g(0)=x_0,g(1)=b$ ($x_0$ is the basepoint and also the common point).
Denote $K\vee T$ by $X$.
The suspension of X can be obtained by a quotient map $q:X\times I\to SX$. Now the basepoint $x_0$ is sent to $x'_0=q(x_0,\frac{1}{2})$. Then Consider two open path connected subspace of $I$ ,they are $(m,1]$ and $[0,n)$ where $m\in(0,1/2),n\in(1/2,1)$, then $ x'_0\in q(X\times (m,1])\cap q(X\times [0,n))$.
Now, let $A=q(X\times (m,1])$ and $B=q(X\times [0,n))$, both of them are contractible because we can slide each point through a path pointing to $SX\times\{1\}$ and $SX\times \{0\}$, respectively. Take $A$ as an example, it can be contract by
$$
G_A((x,s),t)=(x,(1-s)t+s)
$$
A similar construction works for $B$.
So, $\pi_1(A,x'_0)\approx\pi_1(B,x'_0)=0$ and by Seifert-Van Kampen Thm (we can use it because $X$ is path-connected), $ \pi_1(SX,x'_0)$ is trivial.
Best Answer
We have two representations of the Klein bottle as a fundamental polygon. The first is:
And the second is formed by cutting this polygon across the diagonal, flipping one piece and reattaching it to give essentially two real projective planes glued together:
You should be able to see that as CW complexes and a 2-cell attached according to the diagram, these both form the Klein bottle with non-orientable genus 2.
Removing a point from this 2-cell produces a space that deformation rectacts onto the 1-skeleton, which in both cases obviously forms the wedge sum of two circles and the fundamental group is $\Bbb{Z} * \Bbb{Z}$.
Let's see if we can develop some sort of physical intuition for this. If the point (or by deformation, hole) we remove is in the right place, we can embed this in $\Bbb{R}^3$ to get a physical intuition.
Which then forms
And you can see rather easily that this deforms to:
Which obviously has the fundamental group of $\Bbb{Z} * \Bbb{Z}$, as this deformation retracts onto $S^1 \vee S^1$.